How to convert a code from C language to Matlab?

Dear all,

I have a code written in C language that detect the lines between cells in confocal images using the Watershed method. I want to use this C code in matlab. Is there any way to do so?

[mary christeena thomas]

hello Sir,

can you please convert this c code to matlab code and give??

C code:

#include "mex.h"

/* Find variance and mean of the pixels in the window */
void arrayProduct(double v, double In, double *y, double *z, mwSize n)
{
    double VarW,MeanW,W;
    mwSize i;
   
    MeanW=0;
    for (i=0; i
        MeanW=MeanW+y[i];
    }
    MeanW=MeanW/n;
    VarW=0;
    for(i=0;i
    {
        VarW=VarW+((y[i]-MeanW)*(y[i]-MeanW));
    }
    VarW=VarW/n;
   
    W=VarW/(VarW+v);
    z[0]=MeanW+W*(In-MeanW);
}


void mexFunction( int nlhs, mxArray *plhs[],
                  int nrhs, const mxArray *prhs[])
{
    double Rvariance,CIndx,*inMatrix,*out;              
    size_t ncols;               
   
   
    Rvariance = mxGetScalar(prhs[0]);
    CIndx  = mxGetScalar(prhs[1]);
    inMatrix = mxGetPr(prhs[2]);
    ncols = mxGetN(prhs[2]);
   
    plhs[0] = mxCreateDoubleMatrix(1, 1, mxREAL) 
    out = mxGetPr(plhs[0]);

    /* Call the function arrayProduct */
    arrayProduct(Rvariance,CIndx,inMatrix,out,(mwSize)ncols);
}

[Walter Roberson]

Those for loops have invalid syntax in C .

[Hari Priya]

I want to convert my c code into matlab code ...plz convert and give

[Hari Priya]

#include<stdio.h>
#include<math.h>
int main(){
  float x1,y1,x2,y2,vx,vy,len,x3,y3,size;
  vx=x1-x2;
  vy=y1-y2;
  len=sqrt((vx*vx)+(vy*vy));
  x3=vx/((len*size)+x1);
  y3=vy/((len*size)+y1);
  float temp1,temp2,angle;
  temp1=y3-y1;
  temp2=x3-x1;
  angle=atan2(temp1,temp2)+1.57;
  return 0;
  }

[Walter Roberson]

Hari Pryia, that code does not appear to relate to the Question?

That code uses a number of uninitialized variables.

[Hari Priya]

Ok sir but I want some feature extraction code for image processing can you help me ?

[Walter Roberson]

You should open a new Question.

[otike stanley]

Please can somebody help me convert the below Matlab algorithm to C language. I have tried several times using the Matlab Coder without success. It always return an error message, I would't know how to handle.

function theCode
    %prevent scientific notation
    format long
    
    %edit filenames and sheetnames based on your application
    testDataFilename = 'dataTable.xlsx';
    testDataSheetName = 'modelData';
       
    %Function of the imposition of the threshold polynomial
    P = [-12.88 315.53 4900];
    
    %get string representation and function handle to compute data for
    [polyStr,Ft] = polyVector2StrAndFunc(P);

    %read testData from excel file
    testData = readtable(testDataFilename,'Sheet',testDataSheetName);
    [testData,openCircuit,closeCircuit] = evaluateData(Ft,testData);

    %write results to excel file on separate sheetname
    writetable(testData,testDataFilename,'FileType','spreadsheet','Sheet',sprintf('%s_OutputResult',testDataSheetName))

    %visualize result
    plotFitData(P,polyStr,testData,openCircuit,closeCircuit)

    function [polyStr,polyFunc] = polyVector2StrAndFunc(P)
        polDegree = length(P) - 1;
        polyStr = 'F(t) = ';
        funcStr = '@(t)';
        for p = P

            if p < 0
                pm = ' - ';
            else
                pm = ' + ';
            end
            if polDegree > 1
                tDegStr = sprintf('t^{%d}',polDegree);
                fDegStr = sprintf('t .^ %d',polDegree);
            elseif polDegree == 1
                tDegStr = 't';
                fDegStr = tDegStr;
            else
                tDegStr = '';
                fDegStr = '';
            end

            if strcmp(polyStr,'F(t) = ')
                polyStr = sprintf('%s %.4f%s',polyStr,p,tDegStr);
                funcStr = sprintf('%s %f ',funcStr, p);
            else
                polyStr = sprintf('%s %s %.4f%s',polyStr,pm,abs(p),tDegStr);
                funcStr = sprintf('%s %s %f',funcStr, pm, abs(p));

            end
            if ~strcmp(fDegStr,'')
                funcStr = sprintf('%s %s %s',funcStr, ' .* ',fDegStr);    
            end
            %decrement degree to next level
            polDegree = polDegree - 1;
        end
        %create function to use in computing data
%         disp(funcStr)
        disp(polyStr)
        polyFunc = str2func(funcStr);
    end
    function plotFitData(P,polyStr,testData,openCircuit,closeCircuit)

        figure()
        hold on
        legend show
        plot(testData{:,'Time'},polyval(P,testData{:,'Time'}),...
            'DisplayName',polyStr,'Color','k')
        plot(testData{openCircuit,'Time'},testData{openCircuit,'ActivePower'}, '*',...
            'DisplayName','Open Circuit Breaker', 'Color', 'r')
        plot(testData{closeCircuit,'Time'},testData{closeCircuit,'ActivePower'}, 'o',...
            'DisplayName','Close Circuit Breaker', 'Color', 'g')
        hold off
        legend('FontSize',8,'Location','northoutside')
        %visualize Ft over the 24 hours period
        ylabel('Power')
        xlabel('Time')
    end
    function [testData,openCircuit,closeCircuit] = evaluateData(Ft,testData)
        %compute Ft value for time data in userTable
        testData{:,'Ft'} = Ft(testData{:,'Time'});

        %open circuit breaker rows
        openCircuit = testData{:, 'ActivePower'} > testData{:, 'Ft'};
        testData(openCircuit,'REMARK') = {'Open circuit breaker'};

        %close circuit breaker rows
        closeCircuit =  testData{:, 'ActivePower'} <= testData{:, 'Ft'};
        testData(closeCircuit,'REMARK') = {'Close circuit breaker'};
    end
end

[Walter Roberson]

readtable() and writetable() expand to a lot of code -- too much to convert by hand.

Graphics are not supported by MATLAB Coder. There is no C or C++ standard plotting library (there are several competing standards, including DirectX and OpenGL and Metal)

Answer 1

[Azzi Abdelmalek]

Look at mex files

[Meshooo]

Thank you very much but I wonder if there is more straightforward way/software to do that in easy way. Do you think matlab compiler can do that automatically?

[Azzi Abdelmalek]

Matlab compiler allows to create an executable file from your m code; while Matlab coder convert your m code to c or c++ code. Mex files allows to run your c++ or fortran files in Matlab environment, but no m code is generated.

Answer 2

[zhao]

I guess no direct way to do that. I am considering convert matlab code to C. This should be easier, but it is still not good.

[Machine Learning Enthusiast]

What is the procedure for coverting Matlab code to C?

[Walter Roberson]

MATLAB Coder can convert (some) MATLAB code to C.

Answer 3

[Walter Roberson]

If you compile the C to a DLL, you can use loadlibrary()

Note: historically, Unix .o files were (are?) in the same format as .so (shared object) so historically you did not always have to run a link phase, if you had a single .o .

[Meshooo]

OK, thank you very much Walter.

Answer 4

[N/A]

while(true~=1)
    for i=0; i<25; i+1;
        for j=0; j<25 ; j+1;
                 if s[i]=s[j] && i~=j
                 match[k]=s[i];
                 k=k+1;
                     if s[i]~=s[j] && i~=j               
                     true=0;
                     end
                 end
        end
    end
end

how to convert this to mathlab

[Walter Roberson]

flagtrue = 0;
while (flagtrue~=1)
  i = 0;
  while i < 25
    j = 0;
    while j < 25
      s(i+1) = s(j);
      if s(i+1) ~= 0 && i ~= j
        flagtrue = 0;
      end
      j + 1;   %not sure why you bother to add 1 to j and discard the result
    end
    i + 1;     %not sure why you bother to add 1 to i and discard the result
  end
end

This is the equivalent MATLAB. It is an infinite loop, because the original code is an infinite loop (for multiple reasons).

[Hari Priya]

#include<stdio.h>
#include<math.h>
int main()
{ 
float x1,y1,x2,y2,vx,vy,len,x3,y3,size;
 vx=x1-x2;
 vy=y1-y2;
 len=sqrt((vx*vx)+(vy*vy));
 x3=vx/((len*size)+x1);
 y3=vy/((len*size)+y1);
 float temp1,temp2,angle;
 temp1=y3-y1; 
temp2=x3-x1;
 angle=atan2(temp1,temp2)+1.57;
 return 0; 
}

I want to Convert this code into matlab code sir

Answer 5

[dania mohammed]

BEGIN {
     highest_packet_id = 0;
}
{
   action = $1;
   time = $3;
   #from = $3;
   #to = $4;
   type = $35; #aodv relationships,if 5 no infomation
   pktsize = $37;
   #src = $9;
   #dst = $10;
   #seq_no = $11;
   packet_id = $41;

if ( type != "AODV" ) {

     if ( packet_id > highest_packet_id )
           highest_packet_id = packet_id;

if ( start_time[packet_id] == 0 )

   if ( type == "cbr" && action != "d" ) {
        if ( action == "r" ) {
           end_time[packet_id] = time;
        }
     } else {

end_time[packet_id] = -1;
   }
}
}
END {

   for ( packet_id = 0; packet_id <= highest_packet_id; packet_id++ ) {
         start = start_time[packet_id];
         end = end_time[packet_id];
         packet_duration = end - start;

   if ( start < end ) printf("%f %f\n", start, packet_duration);
     }
  }

please anyone help me

I need convert this code to matlab

how I can get it???

[Walter Roberson]

Which AWK version do you need this to conform with?

This code has no direct translation in MATLAB, because this code assumes "standard input", which MATLAB does not have. The MATLAB version would need to be told which file to open.

Are you aware that you can invoke perl from MATLAB? And there is an AWK to Perl convertor: http://perldoc.perl.org/a2p.html

Answer 6

[Hoang Kien]

lease anyone help me I need convert this code to matlab how I can get it???

 #include <stdio.h>
#include <conio.h>
void main()
{
int n, a, c, i, j;
int h, k;
int hs[36], hsc[36], hsd[36];
printf("Bien doi dang Da thuc.\n");
printf("Chuan tac suy rong sang chuan tac suy rong.\n");
printf("Nhap n, a va h: ");
scanf("%d" "%d" "%d", &n, &a, &h);
printf("Nhap c va k: ");
scanf("%d" "%d", &c, &k);
hsd[n]=0;
for(i=0; i<= n; i++){
printf("Nhap vao he so a%d: ",i);
scanf("%d", &hs[i]);}
printf("\n%5s"," ");
for (i=n; i>=0; i--){
printf("%5d ", hs[i]);}
for (j=0;j<n;j++){
for (i=n-1;i>=j;i--){
hsc[j]=c+(j*k);
hsd[i]=(hsc[j]-a+((j-i)*h)); //hsd[i]=(c+(j*k)-a+((j-i)*h));
hs[i]=hs[i]+hs[i+1]*hsd[i]; //hs[i]=hs[i]+hs[i+1]*(c+(j*k)-a+((j-i)*h));}
printf("\n%5s"," ");
for (i=n; i>=j; i--){
printf("%5d ", hsd[i]);}
printf("\n%5d",hsc[j]);
for (i=n; i>=j; i--){
printf("%5d ", hs[i]);}}
printf("\n\nVoi c = %d va k = %d, cac he so cua da thuc moi la:\n",c,k);
for (i=0; i<=n; i++){
printf("%5d ", hs[i]);}
getch();}

[Walter Roberson]

If you have R2017a or later, then:

• remove the #include
• change void main() to function main
• remove each { character, and replace each } by end
• comment out the int statements
• change the // to %
• change for(i=0; i<= n; i++) to for i=0:n
• change for (i=n; i>=j; i--) to for i=n:-1:j
• change each [] subscript to () and add one to each subscript. So for example hsc[j] would be replaced by hsc(j+1)
• in each place where you have two "" strings beside each other with a space between them, change the space to + . For example "%d" "%d" would become "%d"+"%d"
• change the final getch(); to pause()
• MATLAB does not have an scanf that works from standard input. You will need to make decisions as to how you want to handle the input. Is the intent that you are reading from a file, or that you are prompting the user?

[Murthy Dhulipala]

Sir,

Kindly help me to convert the below c code into MATLAB code

# include <stdio.h>
# include<math.h>

void main()
{
FILE *input_I1Q1, *input_I2Q2, *out_RGB;
int i, A, B,n;
float I1[n],I2[n],Q1[n],Q2[n],S0,S1,S2,S3,m,delta,R[n],G[n],B[n];
//float I1[n],I2[n],Q1[n],Q2[n],S0[n],S1[n],S2[n],S3[n],m[n],delta[n],R[n],G[n],B[n];

input_I1Q1=fopen("I1Q1.bin","r");
input_I2Q2=fopen("I2Q2.bin","r");
out_RGB=fopen("RGB.xls", "w");

for(i=0;i<n;i++) 

{
fscanf(I1Q1, "%f %f",&I1,&Q1);
fscan(I2Q2,"%f%f",&I2,&Q2);
A=(I1)**2+(Q1)**2; 
B=(I2)**2+(Q2)**2;
S0=A+B;
S1=A-B;
S2=2*((I1)*(I2)+ (Q1)*(Q2));
S3=2*((Q1)*(I2)-(Q2)*(I1));
m=(sqrt(S1)**2+(S2)**2+(S3)**2)/S0);
delta=atan2(S3,S2);
R[i]=sqrt(S0*m*(1+sin(delta)/2);
G[i]=sqrt(S0*m*(1-sin(delta)/2);
B[i]=sqrt(S0*(1-m));
fprintf(RGB,"%f\t%f\t%f\n",R[i],G[i],B[i]);
}
fclosef(I1Q1);
fclosef(I2Q2);
fclosef(RGB);
}

[Murthy Dhulipala]

n= 20000

[Walter Roberson]

The below code needs to be reviewed, as it does not do exactly the same thing that the C code does.

1. The C expression for m involved taking the square root of S1 and immediately squaring that, and adding to that the squares of S2 and S3. That seems very unlikely to be what is wanted: it is much more likely that you would square S1, S2, and S3, add those squares, and take the square root of that.
2. The C expressions for calculating R and G were each missing ) and so would have failed to compile. I had to guess where the ) went by analogy to the calculation for B.
3. The C expression for writing the output file would fail to compile, or if it compiled would either crash or do a random thing. This is because the fprintf used a file identifier named RGB, which is not a defined variable in the program. The below MATLAB code is therefore different because it actually writes the variables to a file.
4. C does not use ** for exponentation. C has no exponentation operator and uses the pow() function instead. Your code would not have compiled in C.

The C code seems to be intended to write to a file named 'RGB.xls' but it does so as tab delimited text. This is confusing, and is likely to cause problems. I would recommend that you either write it to a true .xls file or that you write it to a .tsv or .txt file. To write it to a true .xls file, use writematrix(RGB, 'RGB.xls') with no further options. To write it to a .tsv or .txt file, change the file extension in the writematrix() call I provided.

There is another difference with the code I provided: I read and output the entire input files, not just the first n points of it.

In future, when you ask people to convert code for you, you should make sure that it is working code, so that they do not have to make guesses about what the code is intended to do.

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

 S = fileread('I1Q1.bin');
 IQ1 = sscanf(S, '%f');
 S = fileread('I2Q2.bin');
 IQ2 = sscanf(S, '%f');
 I1 = IQ1(1:2:end);
 Q1 = IQ2(2:2:end);
 I2 = IQ2(1:2:end);
 Q2 = IQ2(2:2:end);
 A = I1.^2 + Q1.^2;
 B = I2.^2 + Q2.^2;
 S0 = A + B;
 S1 = A - B;
 S2 = 2 * (I1 .* I2 + Q1 .* Q2);
 S3 = 2 * (Q1 .* I2 - Q2 .* I1);
 m = sqrt(S1.^2 + S2.^2 + S3.^2)/S0;   %CHECK THIS!!! (1)
 delta = atan2(S3,S2);
 R = sqrt(S0 .* m .* (1+sin(delta)/2));  %CHECK THIS!!! (2)
 G = sqrt(S0 .* m .* (1-sin(delta)/2));  %CHECK THIS!!! (2)
 B = sqrt(S0*(1-m));
 RGB = [R(:), G(:), B(:)];
 
 writematrix(RGB, 'RGB.xls', 'filetype', 'text', 'delimiter', '\t');  %CHECK THIS!!! (3)
 

Answer 7

[Barenya Bikash Hazarika]

Can anybody convert this C Code to matlab?

#include<stdio.h>
#define MAX_SIZE 5
int main()
{
    int arr[MAX_SIZE]; %Declare an array of MAX_SIZE
    int i,j, N;
      // Input array size 
      printf("Enter size of array: ");
      scanf("%d", &N);
      // Input elements in array 
      printf("Enter %d elements in the array : ", N);
      for(i=0; i<N; i++)
      {
          scanf("%d", &arr[i]);
      }
       // Print all elements of array
      printf("\nElements in array are: ");
      for(i=0; i<N; i++)
      {
          for(j=i; j<N; j++)
          {
          printf("%d, ", arr[i,j]);
          }
          printf("\n");
      }
      return 0;
  }

[Walter Roberson]

No, that C code is broken. It inputs a number for the array size, and then loops that many times reading in one element at a time into elements of arr . But then it tries to go through and print data from arr as if arr were a two-dimensional array.

Like if the user had requested to enter 3 items, then arr[0], arr[1], arr[2] would be stored into, but then the code would request to output arr[0,0], arr[0,1], arr[0,2], arr[1,0], arr[1,1], arr[1,2], arr[2,0], arr[2,1], arr[2,2] . You stored N elements into the array but try to extract N^2 elements from the array.

The code would refuse to compile in C because in order to use arr[i,j] then arr would have had to have been declared as int arr[SOMETHING,SOMETHING] . If you had used arr[i][j] then arr would have had to have been declared int arr[SOMETHING][SOMETHING] which is situation internally involving pointers to subarrays but you only have one vector, not a 2D array and not a vector of pointers to vectors.

[leo john]

Kindly help sir how to do scanf in array using for loop

#include<stdio.h>
#define MAX_SIZE 5
int main()
{
    int arr[MAX_SIZE]; %Declare an array of MAX_SIZE
    int i,j, N;
      // Input array size 
      printf("Enter size of array: ");
      scanf("%d", &N);
      // Input elements in array 
      printf("Enter %d elements in the array : ", N);
      for(i=0; i<N; i++)
      {
          scanf("%d", &arr[i]);
      }
       // Print all elements of array
      printf("\nElements in array are: ");
      for(i=0; i<N; i++)
      {
         
          printf("%d, ", arr[i]);
          
      }
      return 0;
  }

[Walter Roberson]

MATLAB does not have any equivalent to scanf(). scanf() reads from standard input but MATLAB does not have standard input. The closest MATLAB has is to use input() with the 's' option and sscanf() the resulting character vector.

Answer 8

[kavan dave]

 public void BlogECGExample()
    {
      // Define your own dataDir
      var dataDir = "................";

      // Load ECG wave from physionet.org data file.
      string filename = Path.Combine( dataDir, "ECG_AAMIEC13.data.txt" );
      string line;
      int cnt = 0;
      FloatVector ecgMeasurement = new FloatVector( 3000 );
      var fileStrm = new System.IO.StreamReader( filename );
      fileStrm.ReadLine(); fileStrm.ReadLine();
      while ( ( line = fileStrm.ReadLine() ) != null && cnt < 3000 )
      {
        ecgMeasurement[cnt] = Single.Parse( line.Split( ',' )[1] );
        cnt++;
      }

      // Choose wavelet
      var wavelet = new FloatWavelet( Wavelet.Wavelets.D4 );

      // Build DWT object
      var dwt = new FloatDWT( ecgMeasurement.DataBlock.Data, wavelet );

      // Decompose signal with DWT to level 5
      dwt.Decompose( 5 );

      // Find Universal threshold & threshold all detail levels with lambdaU
      double lambdaU = dwt.ComputeThreshold( FloatDWT.ThresholdMethod.Universal, 1 );
      dwt.ThresholdAllLevels( FloatDWT.ThresholdPolicy.Soft, new double[] { lambdaU, lambdaU, lambdaU, lambdaU, lambdaU } );

      // Rebuild the signal to level 1 - the original (filtered) signal.
      float[] reconstructedData = dwt.Reconstruct();

      // Display DWT results.
      BlogECGExampleBuildCharts( dwt, ecgMeasurement, reconstructedData );

    }

    public void BlogECGExampleBuildCharts( FloatDWT dwt, FloatVector ECGMeasurement, float[] ReconstructedData )
    {

      // Plot out approximations at various levels of decomposition.
      var approxAllLevels = new FloatVector();
      for ( int n = 5; n > 0; n-- )
      {
        var approx = new FloatVector( dwt.WaveletCoefficients( DiscreteWaveletTransform.WaveletCoefficientType.Approximation, n ) );
        approxAllLevels.Append( new FloatVector( approx ) );
      }

      var detailsAllLevels = new FloatVector();
      for ( int n = 5; n > 0; n-- )
      {
        var approx = new FloatVector( dwt.WaveletCoefficients( DiscreteWaveletTransform.WaveletCoefficientType.Details, n ) );
        detailsAllLevels.Append( new FloatVector( approx ) );
      }

      // Create and display charts.
      Chart chart0 = NMathChart.ToChart( detailsAllLevels );
      chart0.Titles.Add( "Concatenated DWT Details to Level 5" );
      chart0.ChartAreas[0].AxisY.Title = "DWT Details";
      chart0.Height = 270;
      NMathChart.Show( chart0 );

      Chart chart1 = NMathChart.ToChart( approxAllLevels );
      chart1.Titles.Add("Concatenated DWT Approximations to Level 5");
      chart1.ChartAreas[0].AxisY.Title = "DWT Approximations";
      chart1.Height = 270;
      NMathChart.Show( chart1 );

      Chart chart2 = NMathChart.ToChart( (new FloatVector( ReconstructedData ))[new Slice(500,500)] );
      chart2.Titles[0].Text = "Thresholded & Reconstructed ECG Signal";
      chart2.ChartAreas[0].AxisY.Title = "mV";
      chart2.Height= 270;
      NMathChart.Show( chart2 );

      Chart chart3 = NMathChart.ToChart( (new FloatVector( ECGMeasurement ))[new Slice(500,500)] );
      chart3.Titles[0].Text = "Raw ECG Signal";
      chart3.ChartAreas[0].AxisY.Title = "mV";
      chart3.Height = 270;
      NMathChart.Show( chart3 );

    }

Can you please tell me how to convert this code into matlab code? what to write in place of public void ()

[Walter Roberson]

In MATLAB, the equivalent of

public void BlogECGExample()

is

function BlogECGExample

with the code stored in BlogECGExample.m

Answer 9

[Mahboubeh Parastarfeizabadi]

Can anyone helping me to convert this C code to Matlab:

#include "FFilter.h"

static int filter_taps[FFILTER_TAP_NUM] = {
  158,
  2,
  2,
  1,
  1,
  1,
  0,
  0,
  -1,
  -2,
  -2,
  -3,
  -5,
  -6,
  -7,
  -9,
  -10,
  -12,
  -14,
  -15,
  -18,
  -20,
  -22,
  -24,
  -26,
  -29,
  -31,
  -34,
  -36,
  -39,
  -41,
  -44,
  -46,
  -48,
  -50,
  -52,
  -53,
  -55,
  -56,
  -57,
  -58,
  -58,
  -58,
  -57,
  -57,
  -56,
  -54,
  -52,
  -50,
  -47,
  -44,
  -41,
  -37,
  -33,
  -28,
  -23,
  -18,
  -12,
  -7,
  -1,
  6,
  12,
  18,
  25,
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};

void FFilter_init(FFilter* f) {
  int i;
  for(i = 0; i < FFILTER_TAP_NUM; ++i)
    f->history[i] = 0;
  f->last_index = 0;
}

void FFilter_put(FFilter* f, int input) {
  f->history[f->last_index++] = input;
  if(f->last_index == FFILTER_TAP_NUM)
    f->last_index = 0;
}

int FFilter_get(FFilter* f) {
  long long acc = 0;
  int index = f->last_index, i;
  for(i = 0; i < FFILTER_TAP_NUM; ++i) {
    index = index != 0 ? index-1 : FFILTER_TAP_NUM-1;
    acc += (long long)f->history[index] * filter_taps[i];
  };
  return acc >> 16;
}
#ifndef FFILTER_H_
#define FFILTER_H_

/*

FIR filter designed with
 http://t-filter.appspot.com

sampling frequency: 2000 Hz

fixed point precision: 16 bits

* 0 Hz - 10 Hz
  gain = 0
  desired attenuation = -40 dB
  actual attenuation = n/a

* 13 Hz - 30 Hz
  gain = 1
  desired ripple = 5 dB
  actual ripple = n/a

* 33 Hz - 1000 Hz
  gain = 0
  desired attenuation = -40 dB
  actual attenuation = n/a

*/

#define FFILTER_TAP_NUM 723

typedef struct {
  int history[FFILTER_TAP_NUM];
  unsigned int last_index;
} FFilter;

void FFilter_init(FFilter* f);
void FFilter_put(FFilter* f, int input);
int FFilter_get(FFilter* f);

#endif

Answer 10

[Cristian Macavei]

Can someone help me convert this to matlab please?

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#ifndef M_PI
#define M_PI 3.1415926535897
#endif

int main()
{
    int n;
    float proiectieX[100], proiectieY[100];

    printf("Numar de vectori: ");
    scanf("%d" , &n);

    for(int i = 1; i <= n ; ++i)
    {
        float theta;
        float modul;
        printf("Modulul vectorului %d si unghiul dintre Ox si vector in grade:", i);
        scanf("%f %f", &modul, &theta);

        if(modul < 0) continue;

        theta = (theta * M_PI) / 180.0f;

        proiectieX[i] = cos(theta) * modul;
        proiectieY[i] = sin(theta) * modul;

    }

    for (int i = 1; i <= n; ++i)
    {
        printf("Proiectiile vectorului %d : %g %g \n", i, proiectieX[i], proiectieY[i]);
    }


    return 0;
}

[Walter Roberson]

please start aa new question

[pallavi sapkale]

please convert this c code in matlab...........

/* Includes the library functions of C */
#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include <ctype.h>
#include <string.h>
#include <math.h>
#include <time.h>
/* Declaration of Global Variables for time control of Simulation */
time_t start, end;
/* For scaling the time in order quicked the simulation */
int scale_time = 0;
int scale_measure = 0;
/* Seed for generating random numbers */
long *seed;
/* For keeping count of the changes to different networks and time
spent in each network */
int count_wlan = 0;
int count_cdma = 0;
int wlan_cdma = 0;
/* Opens the file for writing the solution */
FILE *fpt = fopen("SimulationResults.txt", "wb");
60
/* Function for reading data from file */
int readln(FILE *fp, char *ch)
{
char c;
int i = 0;
if (fread(&c, 1, 1, fp)<1)
return 0;
while (c=='\n' || c=='\r' || c==' ')
if (fread(&c, 1, 1, fp)<1)
return 0;
while (!(c=='\n' || c == '\r'))
{
*ch = c;
ch++;
i++;
if (fread(&c, 1, 1, fp)<1)
return 0;
}
*ch=0;
return 1;
}
/* END readln */
/* Random number generator functions */
/******************************************************************
Function: rndnum
Portable uniform (0,1) random number generator. Uses the
multiplicative congruential method:
z(i)=(7**5*z(i-1))(mod 2**31 - 1)
Ref : Law & Kelton, SIMULATION MODELING AND ANALYSIS,
McGraw Hill, 1982, p. 227.
******************************************************************/
double rndnum(
long *seed ) /* long integer seed (by reference) */
{
static long a = 16807;
static long b15 = 32768;
static long b16 = 65536;
static long p = 2147483647;
static long xhi, xalo, leftlo, fhi, k;
static double lawkel;
xhi = *seed/b16;
xalo = ( *seed - xhi * b16) * a;
leftlo = xalo / b16;
fhi = xhi * a + leftlo;
k = fhi / b15;
*seed = ( ( ( xalo - leftlo * b16 ) - p )+( fhi - k * b15 ) * b16 ) +
k;
if ( *seed < 0 ) *seed = *seed + p;
lawkel = *seed * 4.656612875e-10;
return lawkel;
}
61
/******************************************************************
Function: drand
Generate a (continuous) uniform random variable in the
interval [a,b] using a passed seed. Returns the value.
******************************************************************/
double drand(
double a, /* low end of the range */
double b, /* high end of the range */
long *seed ) /* long integer seed (by reference) */
{
double value;
value = (a + (b - a) * rndnum(seed));
return value;
}
/******************************************************************
Function: irand
Generate an integer uniform random variable in the (integer)
interval [a,b] using a passed seed. Returns the value.
******************************************************************/
int irand(
int a, /* low end of the range */
int b, /* high end of the range */
long *seed ) /* long integer seed (by reference) */
{
int trial;
trial = a + (int) ( (b - a + 1) * rndnum(seed) );
if (trial > b)
{
trial = b;
}
return trial;
}
/* Declaration of functions that will be used */
int loop_wlan(double Chi_thresh, int real_time, int lambda_r, int
lambda_n, int Num_RSS, double RSS, int NEARWLAN, int lambda_u, int
Sim_time, int Mea_time);
int loop_cdma(int NEARWLAN, double Chi_thresh, int lambda_u, int Num_RSS,
double RSS, int real_time, int lambda_r, int lambda_n, int Sim_time, int
Mea_time);
int loop_cdma1(int NEARWLAN, double Chi_thresh, int lambda_u, int
Num_RSS, double RSS, int real_time, int lambda_r, int lambda_n, int
Sim_time, int Mea_time);
/* Main starts here */
int main()
{
62
/* Declaration of Variables and Parameters */
double RSS = 0;
double Chi_thresh = 0;
int Num_RSS = 0;
int WORKWLAN = 0;
int NEARWLAN = 0;
int real_time = 0;
int lambda = 0;
int lambda_r = 0;
int lambda_n = 0;
int lambda_u = 0;
int Sim_time = 0;
int Mea_time = 0;

[Walter Roberson]

You should start a new question for this.

Is it necessary to duplicate the same random number generators for compatibility, or could you just use rand() and irand()?

Does the implementation of readln() have to be bug-for-bug compatible, or could you just use fgetl() ?

We will need to know exactly which compiler and compiler release this code is intended for, in order to figure out how to deal with the 62 that is hanging there at the beginning of main(). That is not legal C or C++, but since compilers are permitted to make extensions, maybe it is an instruction to the compiler about linkage properties, or maybe it is a machine code instruction that is to be inserted.

Answer 11

[EROL EREN GÖZLÜ]

#include <math.h>
#include <stdio.h>
/*********************************************************************
*
*
* Name: mag_est.c
*
* Synopsis:
*
*   Demonstrates and tests the "Alpha * Min + Beta * Max" magnitude
*   estimation algorithm.
*
* Description:
*
*   This program demonstrates the "Alpha, Beta" algorithm for 
*   estimating the magnitude of a complex number.  Compared to
*   calculating the magnitude directly using sqrt(I^2 + Q^2), this
*   estimation is very quick.
*
*   Various values of Alpha and Beta can be used to trade among RMS
*   error, peak error, and coefficient complexity.  This program
*   includes a table of the most useful values, and it prints out the
*   resulting RMS and peak errors.
*
* Copyright 1999  Grant R. Griffin
*
*                    The Wide Open License (WOL)
*
* Permission to use, copy, modify, distribute and sell this software
* and its documentation for any purpose is hereby granted without
* fee, provided that the above copyright notice and this license
* appear in all source copies. THIS SOFTWARE IS PROVIDED "AS IS"
* WITHOUT EXPRESS OR IMPLIED WARRANTY OF ANY KIND. See 
* http://scopeplot.com/dg/wol.htm for more information.
*
*********************************************************************/
/********************************************************************/
double alpha_beta_mag(double alpha, double beta, double inphase,
                      double quadrature)
{
   /* magnitude ~= alpha * max(|I|, |Q|) + beta * min(|I|, |Q|) */
   double abs_inphase = fabs(inphase);
   double abs_quadrature = fabs(quadrature);
   if (abs_inphase > abs_quadrature) {
      return alpha * abs_inphase + beta * abs_quadrature;
   } else {
      return alpha * abs_quadrature + beta * abs_inphase;
   }
}
/*********************************************************************/
double decibels(double linear)
{
   #define SMALL 1e-20
   if (linear <= SMALL) {
      linear = SMALL;
   }
   return 20.0 * log10(linear);
}
/*********************************************************************/
void test_alpha_beta(char *name, double alpha, double beta,
                     int num_points)
{
   #define PI 3.141592653589793
   int ii;
   double phase, real, imag, err, avg_err, rms_err;
   double peak_err = 0.0;
   double sum_err = 0.0;
   double sum_err_sqrd = 0.0;
   double delta_phase = (2.0 * PI) / num_points;
   for (ii = 0; ii < num_points; ii++) {
      phase = delta_phase * ii;
      real = cos(phase);
      imag = sin(phase);
      err = sqrt(real * real + imag * imag)
            - alpha_beta_mag(alpha, beta, real, imag);
      sum_err += err;
      sum_err_sqrd += err * err;
      err = fabs(err);
      if (err > peak_err) {
         peak_err = err;
      }
   }
   avg_err = sum_err / num_points;
   rms_err = sqrt(sum_err_sqrd / num_points);
   printf("%-16s %14.12lf %14.12lf  %9.6lf %4.1lf %4.1lf\n",
          name, alpha, beta, avg_err, decibels(rms_err),
          decibels(peak_err));
}
/*********************************************************************/
void main(void)
{
   #define NUM_CHECK_POINTS 100000
   typedef struct tagALPHA_BETA {
      char *name;
      double alpha;
      double beta;
   } ALPHA_BETA;
   #define NUM_ALPHA_BETA 16
   const ALPHA_BETA coeff[NUM_ALPHA_BETA] = {
      { "Min RMS Err",      0.947543636291, 0.3924854250920 },
      { "Min Peak Err",     0.960433870103, 0.3978247347593 },
      { "Min RMS w/ Avg=0", 0.948059448969, 0.3926990816987 }, 
      { "1, Min RMS Err",              1.0,     0.323260990 },
      { "1, Min Peak Err",             1.0,     0.335982538 },
      { "1, 1/2",                      1.0,      1.0 / 2.0  },
      { "1, 1/4",                      1.0,      1.0 / 4.0  },
      { "Frerking",                    1.0,            0.4  },
      { "1, 11/32",                    1.0,     11.0 / 32.0 },
      { "1, 3/8",                      1.0,      3.0 / 8.0  },
      { "15/16, 15/32",        15.0 / 16.0,     15.0 / 32.0 },
      { "15/16, 1/2",          15.0 / 16.0,      1.0 / 2.0  },
      { "31/32, 11/32",        31.0 / 32.0,     11.0 / 32.0 },
      { "31/32, 3/8",          31.0 / 32.0,      3.0 / 8.0  },
      { "61/64, 3/8",          61.0 / 64.0,      3.0 / 8.0  },
      { "61/64, 13/32",        61.0 / 64.0,     13.0 / 32.0 }
   };
   int ii;
   printf("\n             Alpha * Max + Beta * Min Magnitude Estimator\n\n");
   printf("Name                  Alpha           Beta       Avg Err  RMS   Peak\n");
   printf("                                                 (linear) (dB)  (dB)\n");
  
printf("---------------------------------------------------------------------\n");
   for (ii = 0; ii < NUM_ALPHA_BETA; ii++) {
      test_alpha_beta(coeff[ii].name, coeff[ii].alpha, coeff[ii].beta, 1024);
   }
}

Can anyone convert this code into matlab please?

[Walter Roberson]

You should start a new Question for that. When you do, I recommend that you post your attempt at the conversion, and post particular questions you have such as error messages you encountered.

Answer 12

[MADHUMITHA M]

can we able to convert this into MATLAB

#include <stdint.h>
static inline uint64_t rotl(const uint64_t x, int k)
 {
        return (x << k) | (x >> (64 - k));
} 
static uint64_t s[2];
 
uint64_t next(void) {
        const uint64_t s0 = s[0];
        uint64_t s1 = s[1];
        const uint64_t result = s0 + s1;
 
        s1 ^= s0;
        s[0] = rotl(s0, 24) ^ s1 ^ (s1 << 16); // a, b
        s[1] = rotl(s1, 37); // c
 
        return result;
}
 void jump(void) {
        static const uint64_t JUMP[] = { 0xdf900294d8f554a5, 0x170865df4b3201fc };
 
        uint64_t s0 = 0;
        uint64_t s1 = 0;
        for(int i = 0; i < sizeof JUMP / sizeof *JUMP; i++)
               for(int b = 0; b < 64; b++) {
                       if (JUMP[i] & UINT64_C(1) << b) {
                               s0 ^= s[0];
                               s1 ^= s[1];
                       }
                       next();
               }
 
        s[0] = s0;
        s[1] = s1;
}
/* This is the long-jump function for the generator. It is equivalent to
   2^96 calls to next(); it can be used to generate 2^32 starting points,
   from each of which jump() will generate 2^32 non-overlapping
   subsequences for parallel distributed computations. */
 
void long_jump(void) {
        static const uint64_t LONG_JUMP[] = { 0xd2a98b26625eee7b, 0xdddf9b1090aa7ac1 };
 
        uint64_t s0 = 0;
        uint64_t s1 = 0;
        for(int i = 0; i < sizeof LONG_JUMP / sizeof *LONG_JUMP; i++)
               for(int b = 0; b < 64; b++) {
                       if (LONG_JUMP[i] & UINT64_C(1) << b) {
                               s0 ^= s[0];
                               s1 ^= s[1];
                       }
                       next();
               }
 
        s[0] = s0;
        s[1] = s1;
}
 

[Walter Roberson]

bitshift(), bitor() .

sscanf('0xd2a98b26625eee7b', '%lx')

[Steven Lord]

Is there a reason you're trying to build your own random number generator rather than using the ones included in MATLAB? Several of the existing generators have multiple stream and substream support.

[MADHUMITHA M]

Thank you for your answers.

I'm trying to work on generating this code in the matlab

[Walter Roberson]

That does not really answer the question about why you want to use that specific random number generator.

[MADHUMITHA M]

I need to combine these codes with my datum so i'm searching for the matlab version of this code

[MADHUMITHA M]

And also thanks for replying me.

[Walter Roberson]

To use the below, call myrand() and assign the result to a variable. The result will be a struct with three fields, "next", "jump", and "long_jump", each a function handle. Invoke the function handles out of the struct. For example,

mr = myrand();
val = mr.next();
mr.long_jump();

The below code is untested.

%the "myrand" prefix was used in the below code to avoid naming a function "next"
%to avoid classes with MATLAB's next loop control.

function funs = myrand()
    funs.next = @myrand_next;
    funs.jump = @myrand_jump;
    funs.long_jump = @myrand_long_jump;
    
    persistent s;
    s = zeros(1,2,'uint64');
    
    function result = myrand_rotl(x, k)
        assert( isa(x, 'uint64') )
        result = bitor( x .* 2.^k, x ./ 2.^(64 - k));
    end
    
    function result = myrand_leftshift(x, k)
        %C++ << operator loses bits at the top, but MATLAB saturates
        assert( isa(x, 'uint64') )
        mask = intmax('uint64')./(2.^k) - 1;
        result = bitand( x, mask ) * 2.^k;
    end
    
    function result = myrand_next()
        s0 = s(1);
        s1 = s(2);
        result = s0 + s1;
        
        s1 = bitxor(s1, s0);
        s(1) = bitxor(bitxor(myrand_rotl(s0, 24), s1), myrand_leftshift(s1, 16)); %// a, b
        s(2) = myrand_rotl(s1, 37); %// c
    end
    
    function myrand_jump()
        JUMP = sscanf('0xdf900294d8f554a5 0x170865df4b3201fc', '%lx %lx');
        
        s0 = uint64(0);
        s1 = uint64(0);
        for i = 1:length(JUMP)
            for b = 0 : 63
                if bitget(JUMP(i), b+1)
                    s0 = bitxor(s0, s(1));
                    s1 = bitxor(s1, s(2));
                end
                myrand_next();
            end
        end
        s(1) = s0;
        s(2) = s1;
    end
    
    %{
    /* This is the long-jump function for the generator. It is equivalent to
       2^96 calls to next(); it can be used to generate 2^32 starting points,
       from each of which jump() will generate 2^32 non-overlapping
       subsequences for parallel distributed computations. */
    %}
    
    function myrand_long_jump
        LONG_JUMP = sscanf('d2a98b26625eee7b dddf9b1090aa7ac1', '%lx %lx');
        
        s0 = uint64(0);
        s1 = uint64(1);
        for i = 1 : length(LONG_JUMP)
            for b = 0 : 63
                
                if bitget(LONG_JUMP(i), b+1)
                    s0 = bitxor(s0, s(1));
                    s1 = bitxor(s1, s(2));
                end
                myrand_next();
            end
        end
        s(1) = s0;
        s(2) = s1;
    end
end

[Madhumitha Munusamy]

Thank you so much sir.

I will run the code and let you know the output

Answer 13

[RUBEN NAHUE]

HELLO PLEASE COULD HELP ME HOW TO CONVERT THIS CODE TO MATLAB.

function[L1;L2;U1;U2] = Cholesky(a; b; c)
n = length(a);
Diagonales = zeros(1; n);
Sub = zeros(1; n - 1);
if a(1) > 0 then
Diagonales(1) = sqrt(a(1));
Sub(1) = b(1)/Diagonales(1);
k = 2;
fin = 0;
while k <= n ^ fin == 0 do
if a(k) - Sub(k - 1)^2 > 0 then
Diagonales(k) = sqrt(a(k) - Sub(k -1)^2);
if Diagonales(k) = 0 ^ k < n then
Sub(k) = b(k)/Diagonales(k);
else if Diagonales(k) == 0 then
fprintf('Division por cero')
fin = 1;
end if
else
fprintf('Division por cero')
fin = 1;
end if
k = k + 1
end while
fprintf('Valor negativo en la posicion (1), no se puede ejecutar el metodo')
end if

[Walter Roberson]

That is not C code. It is a language similar to Maple but it is not Maple.

[Walter Roberson]

This is an incorrect transcription of algorithm 2 from http://www.eafit.edu.co/programas-academicos/pregrados/ingenieria-matematica/practicas-investigativas/Documents/sistemas-ecuacines-lineales-dispersas.pdf

If you were to use the original and change the ^ in the if tests to & then it might possibly work as Octave code.

Answer 14

[Madhumitha Munusamy]

i am having a little problem over here. can someone help me

i have a OFDM matlab code. In that i want to replace cyclic prefix by NPR filter to analyze the performance but it's showing error.

below i have given both my matlab code and filter

Answer 15

[NIPUNA RAVEENDRAN]

Can anybody convert this C Code to matlab?

//   2D HYBRID MEDIAN FILTER implementation
//     image  - input image
//     result - output image
//     N      - width of the image
//     M      - height of the image
void _hybridmedianfilter(const element* image, element* result, int N, int M)
{
   //   Move window through all elements of the image
   for (int m = 1; m < M - 1; ++m)
      for (int n = 1; n < N - 1; ++n)
      {
         element window[5];
         element results[3];
         //   Pick up cross-window elements
         window[0] = image[(m - 1) * N + n];
         window[1] = image[m * N + n - 1];
         window[2] = image[m * N + n];
         window[3] = image[m * N + n + 1];
         window[4] = image[(m + 1) * N + n];
         //   Get median
         results[0] = median(window, 5);
         //   Pick up x-window elements
         window[0] = image[(m - 1) * N + n - 1];
         window[1] = image[(m - 1) * N + n + 1];
         window[2] = image[m * N + n];
         window[3] = image[(m + 1) * N + n - 1];
         window[4] = image[(m + 1) * N + n + 1];
         //   Get median
         results[1] = median(window, 5);
         //   Pick up leading element
         results[2] = image[m * N + n];
         //   Get result
         result[(m - 1) * (N - 2) + n - 1] = median(results, 3);
      }
}

[Walter Roberson]

%//   2D HYBRID MEDIAN FILTER implementation
%//     image  - input image
%//     result - output image
%//     N      - width of the image
%//     M      - height of the image
function result = hybridmedianfilter(image, N, M)

   %//   Move window through all elements of the image
   for m = 1: M-2
      for n = 1: N-2
         %   Pick up cross-window elements
         window(0+1) = image((m - 1) * N + n + 1);
         window(1+1) = image(m * N + n - 1 + 1);
         window(2+1) = image(m * N + n + 1);
         window(3+1) = image(m * N + n + 1 + 1);
         window(4+1) = image((m + 1) * N + n + 1);
         %//   Get median
         results(0+1) = median(window);
         %//   Pick up x-window elements
         window(0+1) = image((m - 1) * N + n - 1 + 1);
         window(1+1) = image((m - 1) * N + n + 1 + 1);
         window(2+1) = image(m * N + n + 1);
         window(3+1) = image((m + 1) * N + n - 1 + 1);
         window(4+1) = image((m + 1) * N + n + 1 + 1);
         %//   Get median
         results(1+1) = median(window);
         %//   Pick up leading element
         results(2+1) = image(m * N + n + 1);
         %//   Get result
         result((m - 1) * (N - 2) + n - 1 + 1) = median(results);
      end
   end

Note that median() is not part of C or C++. This matters because some common implementations of median() do not use the same order or meaning of parameters, so we have to guess about which implementation you are using and what the meaning of the parameters is.

In practice no-one would write code this way.

In practice there is a significant chance that this code will not do what you want unless you import a whole bunch of other code that stores arrays in the same order as is used here.

Answer 16

[Josias Sandoval]

Can anybody convert this C Code to matlab please?

#include <stdio.h>
#include <stdlib.h>
FILE *f1,*f2;
int Vth=10,s=0,d=0;
void aztec (int Vth) { /* Entrada, thresold */
enum {PLATEAU,SLOPE};
long Vmx,Vmn,Vmxi,Vmni;
int V,
LineMode=PLATEAU, /* Indica modo Plateau o Slope */
LineLen=1,
V1,
T1,
Vsi,
Tsi,
Sign,
AZT;
int count=0,np=0,ns=0,sp=0,ss=0;
fscanf (f1,"%x\n",&V);
count++;
Vmni=Vmxi=V;
AZT=V;
LineMode=PLATEAU;
while (!feof (f1)) {
while (LineLen<51) {
LineLen++;
if (feof (f1)) break;
fscanf (f1,"%x\n",&V);
count++;
Vmx=Vmxi;
Vmn=Vmni;
if (Vmx<V) Vmxi=V;
if (Vmn>V) Vmni=V;
if ((Vmxi-Vmni)>=Vth) break;
}
T1=LineLen-1;
V1=(Vmx+Vmn)>>1;
if (LineMode==PLATEAU) {
if (T1>2) {
fprintf (f2,"%04x\n",T1);
fprintf (f2,"%04x\n",V1);
sp+=T1; np++;
AZT=V1;
}
else {
LineMode=SLOPE;
if ((V1-AZT)<0) Sign=-1;
else Sign=1;
Tsi=T1;
Vsi=V1;
}
}
else {
if (T1>2) {
fprintf (f2,"%04x\n",-Tsi);
fprintf (f2,"%04x\n",Vsi);
fprintf (f2,"%04x\n",T1);
fprintf (f2,"%04x\n",V1);
sp+=T1; np++;
ss+=Tsi; ns++;
AZT=V1;
LineMode=PLATEAU;
}
else {
if (((V1-Vsi)*Sign)<0) {
fprintf (f2,"%04x\n",-Tsi);
fprintf (f2,"%04x\n",Vsi);
ss+=Tsi; ns++;
AZT=Vsi;
Tsi=T1;
Vsi=V1;
Sign*=-1;
}
else {
Tsi+=T1;
Vsi=V1;
}
}
}
Vmxi=Vmni=V;
LineLen=1;
}
printf ("\nNúmero de muestras tratadas: %d",count);
printf ("\nNúmero de líneas generadas : %5d\tsuma: %d",np,sp);
printf ("\nNúmero de rampas generadas : %5d\tsuma: %d",ns,ss);
}

[Walter Roberson]

What is the format of the input file it is processing? I could work it out from the code, but it is easier to have you describe it.

[Josias Sandoval]

I don't know but I found the code in this pdf page 96 of the book

[Walter Roberson]

When I go through that code, it appears to me as if the code is designed to fetch its input from the C25 and send the input to somewhere I cannot tell at the moment. But it is not clear at all from the code; it just relies upon two files already being open (or connected).

Is your purpose to talk to a C25 to do real-time EEG compression? As that is what the whole premise is, real-time compression. Or do you have files of EEG input that you want to apply their AZTEC algorithm to? If so then what is the input format?

I do not want to go through the trouble of loading the development system to work through the meaning of some of the less clear parts of the code. The device was current more than 20 years ago; I doubt I could run the development system on my Mac, and I do not have one of the devices and I do not have the peripherals needed to test with.

I would suggest that you use the ideas described, and the block diagram in order to write the code from scratch, for use on EEG files that you already have. I can see from the code that the AZTEC algorithm is intended for use only on a single input channel, so you might also want to enhance it to handle multiple channels.

Answer 17

[Bayu Setiaji]

Dear all,

I have a code written in C language that detect the lines between cells in confocal images using the Watershed method. I want to use this C code in matlab. Is there any way to do so?

static void Main(string[] args)
        {
            Random random = new Random();
            int counter = 0;
            double deviasi = 50;
            double mean = 1000;
            double[] data = { 0.0, 0.0 };

            while (counter < 2) {
                
                double rand = random.NextDouble();
                double y = deviasi * rand + mean;
                if (y >= 999.5) {
                    if (y <= 1000.5) {
                        data[counter] = y;
                        counter++; 
                    }
                }            
            }
            Console.WriteLine(data[0]+" "+ data[1]);
            Console.ReadKey();
        }

[Walter Roberson]

counter = 1;
deviasi = 50;
mean = 1000;
data = [0. 0.];
while counter <= 2
    y = deviasi * rand + mean;
    if y >= 999.5 && y <= 1000.5
        data(counter) = y;
        counter = counter + 1;
    end
end
fprintf('%g %g\n', data);
pause()

Answer 18

[Ta Huu Tai]

You can fix it for me??

clear;
clc;
delete(instrfind);
figure;
for theta=0:1:180
for i=1:1:50
    x1(i)=i*cos(theta*3.14/180);
    y1(i)=i*sin(theta*3.14/180);
end
plot(0,0,x1,y1,'color','g','LineWidth',10);
hold on;
end
xlabel('Distance (cm)');
for R=10:10:50
k=1;
for i=0:1:180
    x(k)=R*cos(i*3.14/180);
    y(k)=R*sin(i*3.14/180);
k=k+1;
end
plot(x,y,'color','k','LineWidth',4);
hold on;
end
for theta=0:30:180
for i=1:1:50
    x11(i)=i*cos(theta*3.14/180);
    y11(i)=i*sin(theta*3.14/180);
end
plot(0,0,x11,y11,'color','k','LineWidth',4);
hold on;
end
labels=cellstr(num2str([0:30:180]'));
text([50*cos(0);50*cos(3.18/6);50*cos(2*3.18/6);50*cos(3*3.18/6);50*cos(4*3.18/6);50*cos(5*3.18/6);50*cos(3.18)],[50*sin(0);50*sin(3.18/6);50*sin(2*3.18/6);50*sin(3*3.18/6);50*sin(4*3.18/6);50*sin(5*3.18/6);50*sin(3.18)],labels,'BackgroundColor',[.7 .9 .7]);
s=serial('COM10');
s.BaudRate=9600;
fopen(s);
while(1)
if (fscanf(s,'%f')=='.')==[1 0 0]
    angle=fscanf(s,'%f');
    distance=fscanf(s,'%f');
 elseif (fscanf(s,'%f')=='.')==[1 0 0]
    angle=fscanf(s,'%f');
    distance=fscanf(s,'%f');
else
    fscanf(s,'%f');
    angle=fscanf(s,'%f');
    distance=fscanf(s,'%f');
end
if distance>50
distance=50;
end
x3=distance*cos(angle*3.14/180);
y3=distance*sin(angle*3.14/180);
x4=50*cos(angle*3.14/180);
y4=50*sin(angle*3.14/180);
x5=distance*cos((angle+2)*3.14/180);
y5=distance*sin((angle+2)*3.14/180);
x6=50*cos((angle+2)*3.14/180);
y6=50*sin((angle+2)*3.14/180);
x7=distance*cos((angle-2)*3.14/180);
y7=distance*sin((angle-2)*3.14/180);
x8=50*cos((angle-2)*3.14/180);
y8=50*sin((angle-2)*3.14/180);
x9=distance*cos((angle+3)*3.14/180);
y9=distance*sin((angle+3)*3.14/180);
x10=50*cos((angle+3)*3.14/180);
y10=50*sin((angle+3)*3.14/180);
x11=distance*cos((angle-3)*3.14/180);
y11=distance*sin((angle-3)*3.14/180);
x12=50*cos((angle-3)*3.14/180);
y12=50*sin((angle-3)*3.14/180);
X3=distance*cos(angle*3.14/180);
Y3=distance*sin(angle*3.14/180);
X4=50*cos(angle*3.14/180);
Y4=50*sin(angle*3.14/180);
X5=0;
Y5=0;
X6=50*cos((angle+2)*3.14/180);
Y6=50*sin((angle+2)*3.14/180);
X7=0;
Y7=0;
X8=50*cos((angle-2)*3.14/180);
Y8=50*sin((angle-2)*3.14/180);
X9=0;
Y9=0;
X10=50*cos((angle+3)*3.14/180);
Y10=50*sin((angle+3)*3.14/180);
X11=0;
Y11=0;
X12=50*cos((angle-3)*3.14/180);
Y12=50*sin((angle-3)*3.14/180);
hold on;
n=plot([X3,X4,X5,X6,X7,X8,X9,X10,X11,X12],[Y3,Y4,Y5,Y6,Y7,Y8,Y9,Y10,Y11,Y12],'color','y','LineWidth',20);
m=plot([x3,x4,x5,x6,x7,x8,x9,x10,x11,x12],[y3,y4,y5,y6,y7,y8,y9,y10,y11,y12],'color','r','LineWidth',20);
drawnow;
delete(m,n);
end  
fclose(s);

Matlab said :

Warning: Unsuccessful read: Matching failure in format..

Warning: Unsuccessful read: A timeout occurred before the Terminator was reached..

Warning: Unsuccessful read: A timeout occurred before the Terminator was reached..

Warning: Unsuccessful read: A timeout occurred before the Terminator was reached..

Warning: File 'm,n' not found.

> In Untitled (line 94)

Warning: Unsuccessful read: A timeout occurred before the Terminator was reached..

Error using ==

Matrix dimensions must agree.

Error in Untitled (line 38)

if (fscanf(s,'%f')=='.')==[1 0 0]

[Walter Roberson]

delete(m,n);

should be

delete([m;n]);

[Walter Roberson]

if (fscanf(s,'%f')=='.')==[1 0 0]

Caution: fscanf() on a serial port by default reads until the first terminator is encountered or the end of the input buffer is reached. In a number of places in your code, you are assuming that fscanf(s,'%f') returns a particular size of input. If you have reason to expect a particular number of inputs on one line, I recommend that you use fgetl() to fetch a line, and then sscanf() the line to read the items. I recommend that you continually cross-check that the number of items you receive is the same as the number you expect, and that you have active error handling.

When you work with serial ports, it is common for the communications stream to get out of synchronization with what you expect.

Answer 19

[Mehmet]

Hello all, I have code written in C language and I need to convert it to matlab. Can anyone help me for that?

#include <stdio.h>
#include <math.h>
#define PI 3.14

void calculate(int r1, int r2, int h);

int main()
{
	int r1, r2, h;
	
	printf("Enter a r1 value: ");
	scanf("%d", &r1);
	printf("Enter a r2 value: ");
	scanf("%d", &r2);

	for(h = 1; h < 10; h++){
		calculate(r1, r2, h);
	}

	return 0;
}

void calculate(int r1, int r2, int h)
{
	double volume, surfaceArea;
	volume = PI * h * (r1 + r2) / 3;

	printf("Volume for %d is %f \n", h, volume);

	surfaceArea =  PI * (r1-2) * sqrt((pow((r2 + r1), 2)) + pow(h, 2)) + PI * pow(r1, 2);

	printf("Surface area for %d is %f", h, surfaceArea);
}

[Walter Roberson]

function calculate(r1, r2, h)
  PI = 3.14;
  pow = @power;
  printf = @fprintf;
  
  volume = PI * h * (r1 + r2) / 3;
  printf("Volume for %d is %f \n", h, volume);
  surfaceArea =  PI * (r1-2) * sqrt((pow((r2 + r1), 2)) + pow(h, 2)) + PI * pow(r1, 2);
  printf("Surface area for %d is %f", h, surfaceArea);

end

[Image Analyst]

Or you can use pi (lower case) which is a built-in constant, rather than create PI. And other simplifications.

function calculate(r1, r2, h)
  volume = pi * h * (r1 + r2) / 3;
  fprintf('For r1 = %d, r2 = %d, and h = %d\n    Volume = %f.\n', ...
      r1, r2, h, volume);
  surfaceArea =  pi * (r1-2) * sqrt((r2 + r1).^2 + h^2) + pi * r1^2; % Maybe should be r1-r2 instead of r1-2???
  fprintf('    Surface area = %f.\n', surfaceArea);
end

Though, check your formulas. I don't think they're correct. What is this? Is it a cylinder with a circular hole cut out of the axis?

[Walter Roberson]

Using a more accurate value for Pi would make it a different program...

The second part of the code I posted is copied and pasted from the C code; the first part is just some shims to make the second part work.

[Mehmet]

thank you all very much. I also think the formula is wrong, but it was written like this on the paper.

Answer 20

[Ali Abdualaziz]

Can anybody convert this C Code to matlab please?

Int I, j, k, m=4, n=2;

Float x(float t;), xs[n*m],p[m],xa[m][n],ya[m][n],f=1000, fs=8000;

X(t)=sin(2*pi*f*t);

For(i=1;i<=m*n;i++)

Xs[i]=x(i/fs);

For(j=1;j<=m;j++)

For(k=1;k<=n;k++)

Xa[j][k]=xs[j*k];

For(j=1;j<=m;j++)

P[j]=scanf(“enter %d th slice”,j);

For(j=1;j<=m;j++)

For(k=1;k<=n;k++)

Ya[j][k]=xa[p[j]][k];

[Walter Roberson]

That is not valid C code.

[Walter Roberson]

See the guideance I gave at https://www.mathworks.com/matlabcentral/answers/91303-how-to-convert-a-code-from-c-language-to-matlab#comment_544264

[Walter Roberson]

Float x(float t;)

If we assume that somehow the first character of each line got mangled in posting and should be lowercase instead of uppercase, then

float x(float t;)

However, in C, there cannot be any semi-colon inside a type declaration.

If we assume further that the semi-colon was a typing mistake and the real code was

float x(float t)

then that would declare that you want to use an external function named x that accepts a float parameter and returns a float parameter.

X(t)=sin(2*pi*f*t);

Assuming again that the first character has been accidentally upper-cased,

x(t)=sin(2*pi*f*t);

That would be invalid. You declared x as being a external function, but you cannot assign to functions in C. If you wanted to define x then you would have had to use

float x(float t) {
   return sin(2*pi*f*t);
}

and that would have to be separated from the type declarations of the line

Float x(float t;), xs[n*m],p[m],xa[m][n],ya[m][n],f=1000, fs=8000;

Answer 21

[Maria Imdad]

Hi,

This is c Code for "Tiny Encryption Algorithm" can some one please help me to convert it into matlab..

I will be very thankful.

[code deleted -- WDR]

[Walter Roberson]

Sorry, Maria Imdad:

For legal reasons we are not permitted to discuss encryption here.

Answer 22

[Tala sh]

Hi, please can somebody help me to convert this code from C language to Matlab?

///function for encoding textfile
int RleEncodeFile(FILE *inFile, FILE *outFile)
{
    int currChar;                       /* current characters */
    int prevChar;                       /* previous characters */
    unsigned char count;                /* number of characters in a run */


    /* validate input and output files */
    if ((NULL == inFile) || (NULL == outFile))
    {
        errno = ENOENT;
        return -1;
    }


    /* encode inFile */
    prevChar = EOF;     /* force next char to be different */
    count = 0;


    /* read input until there's nothing left */
    while ((currChar = fgetc(inFile)) != EOF)
    {
        fputc(currChar, outFile);


        /* check for run */
        if (currChar == prevChar)
        {
            /* we have a run.  count run length */
            count = 0;


            while ((currChar = fgetc(inFile)) != EOF)
            {
                if (currChar == prevChar)
                {
                    count++;


                    if (count == UCHAR_MAX)
                    {
                        /* count is as long as it can get */
                        fputc(count, outFile);


                        /* force next char to be different */
                        prevChar = EOF;
                        break;
                    }
                }
                else
                {
                    /* run ended */
                    fputc(count, outFile);
                    fputc(currChar, outFile);
                    prevChar = currChar;
                    break;
                }
            }
        }
        else
        {
            /* no run */
            prevChar = currChar;
        }


        if (currChar == EOF)
        {
            /* run ended because of EOF */
            fputc(count, outFile);
            break;
        }
    }


    return 0;
}

///function  for decoding text file 
int RleDecodeFile(FILE *inFile, FILE *outFile)
{
    int currChar;                       /* current characters */
    int prevChar;                       /* previous characters */
    unsigned char count;                /* number of characters in a run */


    /* validate input and output files */
    if ((NULL == inFile) || (NULL == outFile))
    {
        errno = ENOENT;
        return -1;
    }


    /* decode inFile */
    prevChar = EOF;     /* force next char to be different */


    /* read input until there's nothing left */
    while ((currChar = fgetc(inFile)) != EOF)
    {
        fputc(currChar, outFile);


        /* check for run */
        if (currChar == prevChar)
        {
            /* we have a run.  write it out. */
            count = fgetc(inFile);
            while (count > 0)
            {
                fputc(currChar, outFile);
                count--;
            }


            prevChar = EOF;     /* force next char to be different */
        }
        else
        {
            /* no run */
            prevChar = currChar;
        }
    }


    return 0;
}

[Walter Roberson]

Is it necessary for it to duplicate that exactly? Because there are several Run Length Encoding functions already in the File Exchange.

[Tala sh]

@Walter Roberson

I couldnt find what i realy need , I'am searching for code in matlab to encode and decode text file using Run Length Encoding but Unfortunately I did not find..

[Tala sh]

@Walter Roberson

I hope you can help me because I don't have much time to keep searching ,and I will really thank you..

[Walter Roberson]

https://www.mathworks.com/matlabcentral/fileexchange/4955-rle-de-encoding

https://www.mathworks.com/matlabcentral/fileexchange/19559-run-length-encoding

https://www.mathworks.com/matlabcentral/fileexchange/41813-runlength

https://www.mathworks.com/matlabcentral/fileexchange/33166-run-length-encoding-and-decoding

https://www.mathworks.com/matlabcentral/fileexchange/31123-rle-run-length-encoding

https://www.mathworks.com/matlabcentral/fileexchange/46865-run-length-encoding

and others

[Tala sh]

@Walter Roberson thanx alot ,but all of above didn't help me ,

i need to compress (text file), it means that i need to read from text file and execute Run length Encodin on this file so the output will be compressed text file not string or image or vector

[Walter Roberson]

Ths is untested, but I think it should be bug-for-bug compatible with the C source code for encoding. Which has at least three different major bugs in it.

Note: when you ask us to translate C source code for you instead of helping you perform the desired task in MATLAB, then you are likely to get back as close as possible to exact translation of the C code, not matter how warped that makes the MATLAB code, and with all the bugs of the C code deliberately left in. "It is not for us to reason why" you want to use broken C code as the basis for your MATLAB function.

function [status, errno] = RleEncodeFile(inFile, outFile)
    ENONENT = uint32(3025);

    errno = uint32(0);
    status = int32(0);

    %C variables have random values until they are written to

    currChar = randi([-2^31, 2^31-1], 1, 1, 'int32'); %/* current characters */
    prevChar = randi([-2^31, 2^31-1], 1, 1, 'int32'); %/* previous characters */
    count = char(randi([0 255], 1, 1, 'uint8'));      %/* number of characters in a run */

    %/* validate input and output files */
    if inFile < 0 || outFile < 0
        errno(1) = ENOENT;
        status(1) = -1;
        return
    end

    %/* encode inFile */
    prevChar(1) = EOF();     /* force next char to be different */
    count(1) = 0;


    %/* read input until there's nothing left */
    currChar(1) = fgetc(inFile);
    while currChar ~= EOF()
       fputc(currChar, outFile);

        %/* check for run */
        if (currChar == prevChar)
            %/* we have a run.  count run length */
            count(1) = 0;

            currChar(1) = fgetc(inFile);
            while currChar ~= EOF()
                if (currChar == prevChar)
                    count = count + 1;
                    if (count == UCHAR_MAX())
                        %/* count is as long as it can get */
                        fputc(count, outFile);

                        %/* force next char to be different */
                        prevChar = EOF();
                        break;
                    end
                else
                    % /* run ended */
                    fputc(count, outFile);
                    fputc(currChar, outFile);
                    prevChar = currChar;
                    break;
                end
                currChar(1) = fgetc(inFile); 
            end
            currChar(1) = fgetc(inFile);
        else
            % /* no run */
            prevChar = currChar;
        end


        if (currChar == EOF())
            % /* run ended because of EOF */
            fputc(count, outFile);
            break;
        end
    end


    status(1) = 0;
end

function uchar_max = UCHAR_MAX
   uchar_max = uint8(255);
end

function eof = EOF
   eof = int32(-1);
end

function c = fgetc(FILE)
   %input unsigned char is converted to int but EOF is negative
   % https://pubs.opengroup.org/onlinepubs/009695399/functions/fgetc.html
   c = int32(fread(FILE, 1, '*uint8'));
   if isempty(c); c(1) = EOF(); end
end

function fputc(c, FILE)
    %input int is to be converted to unsigned char
    % https://pubs.opengroup.org/onlinepubs/009695399/functions/fputc.html
    bytes = typecast(c, 'uint8');
    fwrite(FILE, bytes(1), 'uint8');
end

[Tala sh]

@Walter Roberson I'am realy thank you for your help me ..

[Walter Roberson]

@walaa al majzoup

Reminder: that C code does not compress text files; it only thinks it does. But you turned down help for writing run length coding in MATLAB, so you got a MATLAB program that does what the C code does, not a program that does run length encoding in MATLAB.

Answer 23

[ad lyn]

please i need to convert this c code to matlab .

 {
        #region Consts

        /// <summary>
        /// Number of blocks.
        /// </summary>
        const int __blockCount = 128;

        /// <summary>
        /// Right hand x coord of the base rectangle, thus also the left hand x coord of the tail
        /// (pre-determined/computed for 128 blocks).
        /// </summary>
        const double __R = 3.442619855899;

        /// <summary>
        /// Area of each rectangle (pre-determined/computed for 128 blocks).
        /// </summary>
        const double __A = 9.91256303526217e-3;

        /// <summary>
        /// Denominator for __INCR constant. This is the number of distinct values this class is capable
        /// of generating in the interval [0,1], i.e. (2^53)-1 distinct values.
        /// </summary>
        const ulong __MAXINT = (1UL << 53) - 1;

        /// <summary>
        /// Scale factor for converting a ULong with interval [0, 0x1f_ffff_ffff_ffff] to a double with interval [0,1].
        /// </summary>
        const double __INCR = 1.0 / __MAXINT;

        /// <summary>
        /// Binary representation of +1.0 in IEEE 754 double-precision floating-point format.
        /// </summary>
        const ulong __oneBits = 0x3ff0_0000_0000_0000UL;

        #endregion

        #region Static Fields

        // __x[i] and __y[i] describe the top-right position of rectangle i.
        static readonly double[] __x;
        static readonly double[] __y;

        // The proportion of each segment that is entirely within the distribution, expressed as ulong where
        // a value of 0 indicates 0% and 2^53-1 (i.e. 53 binary 1s) 100%. Expressing this as an integer value
        // allows some floating point operations to be replaced with integer operations.
        static readonly ulong[] __xComp;

        // Useful precomputed values.
        // Area A divided by the height of B0. Note. This is *not* the same as __x[i] because the area
        // of B0 is __A minus the area of the distribution tail.
        static readonly double __A_Div_Y0;

        #endregion

        #region Static Initialiser

        static ZigguratGaussian()
        {
            // Initialise rectangle position data.
            // __x[i] and __y[i] describe the top-right position of Box i.

            // Allocate storage. We add one to the length of _x so that we have an entry at __x[__blockCount], this avoids having
            // to do a special case test when sampling from the top box.
            __x = new double[__blockCount + 1];
            __y = new double[__blockCount];

            // Determine top right position of the base rectangle/box (the rectangle with the Gaussian tale attached).
            // We call this Box 0 or B0 for short.
            // Note. x[0] also describes the right-hand edge of B1. (See diagram).
            __x[0] = __R;
            __y[0] = GaussianPdfDenorm(__R);

            // The next box (B1) has a right hand X edge the same as B0.
            // Note. B1's height is the box area divided by its width, hence B1 has a smaller height than B0 because
            // B0's total area includes the attached distribution tail.
            __x[1] = __R;
            __y[1] = __y[0] + (__A / __x[1]);

            // Calc positions of all remaining rectangles.
            for(int i=2; i < __blockCount; i++)
            {
                __x[i] = GaussianPdfDenormInv(__y[i-1]);
                __y[i] = __y[i-1] + (__A / __x[i]);
            }

            // For completeness we define the right-hand edge of a notional box 6 as being zero (a box with no area).
            __x[__blockCount] = 0.0;

            // Useful precomputed values.
            __A_Div_Y0 = __A / __y[0];
            __xComp = new ulong[__blockCount];

            // Special case for base box. __xComp[0] stores the area of B0 as a proportion of __R
            // (recalling that all segments have area __A, but that the base segment is the combination of B0 and the distribution tail).
            // Thus __xComp[0] is the probability that a sample point is within the box part of the segment.
            __xComp[0] = (ulong)(((__R * __y[0]) / __A) * (double)__MAXINT);

            for(int i=1; i < __blockCount-1; i++)
            {
                __xComp[i] = (ulong)((__x[i+1] / __x[i]) * (double)__MAXINT);
            }
            __xComp[__blockCount-1] = 0;  // Shown for completeness.

            // Sanity check. Test that the top edge of the topmost rectangle is at y=1.0.
            // Note. We expect there to be a tiny drift away from 1.0 due to the inexactness of floating
            // point arithmetic.
            Debug.Assert(Math.Abs(1.0 - __y[__blockCount-1]) < 1e-10);
        }

[Walter Roberson]

Converting that C# code to MATLAB is somewhat tricky. It takes cleverness to write the MATLAB code to fail the same way that the C# code would fail.

It seems to me that it would be much easier to use MATLAB's default randn() generator; https://blogs.mathworks.com/cleve/2015/05/18/the-ziggurat-random-normal-generator/

[ad lyn]

thnks for answering me .actually all i need is the initialisation part of the ziggurat method programme on matlab

https://blogs.mathworks.com/cleve/2015/05/18/the-ziggurat-random-normal-generator/

here,i didn't find the part wich i'm searching for