PROBLEM IN sprintf

1 visualización (últimos 30 días)
huda nawaf
huda nawaf el 11 de Oct. de 2011
hi,
in the following ode why k become larger than L when use sprintf, where I got this error when run code:
%%%%%%%%%%%%%
Attempted to access b1(5123); index out of bounds because numel(b1)=5122.
Error in ==> webscop at 9
mat(i,j)=b1(k);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
the code:
f=fopen('ws1.txt','r+');%%%%%%open the original data file
b1=fscanf(f,'%d');
b=sprintf('%d\n', b1);
fclose all
L=length(b1);
k=1;i=1
while k <=L
mat(i,j)=b1(k);
k=k+1;
end
i=i+1;
end
thanks
  3 comentarios
Jan
Jan el 11 de Oct. de 2011
@huda: It is still unclear, what you want to achieve. I guess, a single REPMAT call solves your problem.
huda nawaf
huda nawaf el 11 de Oct. de 2011
sorry , there is missing statement
this is the proper code:
f=fopen('ws1.txt','r+');%%%%%% open the original data file
b=fscanf(f,'%d');
b1=sprintf('%d\n', b);
fclose all
L=length(b1);
k=1;i=1
while k <=L
for j=1:4
mat(i,j)=b1(k);
k=k+1;
end
i=i+1;
end
thanks

Iniciar sesión para comentar.

Respuestas (1)

Walter Roberson
Walter Roberson el 11 de Oct. de 2011
Your "i=i+1" appears to be unneeded, in that it is not inside any loop and the end of your routine is immediately afterwards.
You appear to be using the subscript "j" for mat(i,j) but "j" is not defined in anything you show. As we do not know its value we cannot tell you what is going on. Worse yet we will assume that j has its default value of sqrt(-1) and that you are trying to index an impossible location.
Is "mat" completely uninitialized or is it passed in as part of the missing function header? And there must be a function header or else the "end" would be a syntactic error and you would not be able to run the routine at all.
Real code, please!
  5 comentarios
Walter Roberson
Walter Roberson el 11 de Oct. de 2011
Such long running times are pretty common when you do not preallocate the arrays.
Why not replace your inner loop with
mat(i,1:4) = b1(k:k+3);
k = k + 4;
And having done that, why not replace the entire computation with
mat = reshape(b1(:),4,[]).' ;
This should, of course, lead you to ask, "But what if b1 is not an exact multiple of 4 characters?", but since you do not check that in your original code, we must assume that you have external knowledge that this will never be the case.
I am somewhat mystified that you are storing the character representation of b rather than b itself, but I guess you have your reasons.
huda nawaf
huda nawaf el 12 de Oct. de 2011
thank you very much,
I were not believing that my problem can be solved in instant.
I do appreciate that

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by