To check Scalar or Not

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Rakeshwar Elango
Rakeshwar Elango el 29 de Jun. de 2019
Comentada: Walter Roberson el 18 de Jul. de 2021
I have debugged the code and now the Scalar test for this code is failing although I have given a provition in the 3rd line of the code to check wheather the given inputs are scalar or not.
please help!
function valid = valid_date(year,month,day)
if nargin==3
if fix(month) && isscalar(month) && fix(day) && isscalar(day) && fix(year) && isscalar(year) && isscalar(valid_date) && year>0 && month>0 && day>0
if mod(year,4)==0&&mod(year,100)~=0 || mod(year,400)==0&&mod(year,100)==0
if ismember(month,[1,3,5,7,8,10,12]) && ismember(day,[1:31])
valid=true;
elseif month==2 && ismember(day,[1:29])
valid=true;
elseif ismember(month,[4,6,9,11]) && ismember(day,[1:30])
valid=true;
else
valid=false;
end
else
if ismember(month,[1,3,5,7,8,10,12]) && ismember(day,[1:31])
valid=true;
elseif ismember(month,[4,6,9,11]) && ismember(day,[1:30])
valid=true;
elseif month==2 && ismember(day,[1:28])
valid=true;
else
valid=false;
end
end
else
valid=false;
end
else
valid=false;
end
Assessment result: correctVarious inputs
Assessment result: incorrectNon-scalar
Return false if an input is not scalar...
Assessment result: correctThe last day of every month
Assessment result: correctRandom leap years
Assessment result: correctRandom non-leap years
Assessment result: correctRandom dates
  6 comentarios
Sunney Kumar
Sunney Kumar el 12 de Jun. de 2020
Editada: Sunney Kumar el 12 de Jun. de 2020
funtion valid = valid_date(year, month, day)
[i, j] = size(year);
[k, l] = size(month);
[o, p] = size(day);
if (i == 1 && j ~= 1) || (k == 1 && l ~= 1) || (o == 1 && p ~= 1)
valid = false
else
%rest of the code
end
Try this piece of code to find out about whether input is scalar or not.
Hakan Karapinar
Hakan Karapinar el 17 de Jun. de 2020
if (~isscalar(year) || year < 1 || year ~= fix(year) ) || (~isscalar(month) || month < 1 || month ~= fix(month) ) || (~isscalar(day) || day < 1 || day ~= fix(day) )
valid = false
return ;
check my code out bro.we take same course.:D its worked i passed week 6

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Stephen23
Stephen23 el 29 de Jun. de 2019
Editada: Stephen23 el 29 de Jun. de 2019
The reason is likely because of the order you check the inputs with, e.g.:
fix(month) && isscalar(month)
%^^^^^^^^^ what is this supposed to test for?
% ^^^^^^^^^^^^^^ this must be tested first!
If month is non-scalar, then fix(month) will return a non-scalar value, and then this will throw an error because && does not accept non-scalar input values. If you test isscalar first, then for a non-scalar month it will return false (i.e. a scalar!) and then MATLAB does not evaluate anything that follows the &&. Read the documentation to know more:
Note that it is unclear what your fix(...) code is supposed to achieve, but I doubt that it does anything very useful. In any case, the order should be like this:
isscalar(month) && ??? you need to decide what to put here

Más respuestas (1)

iitm_pnkj
iitm_pnkj el 18 de Jul. de 2021
function valid = valid_date(year, month, day)
m31 = [1 3 5 7 8 10 12];
m30 = [4 6 9 11];
if isscalar(year) == 1 && isscalar(month) == 1 && isscalar(day) == 1 && year > 0 && month > 0 && day > 0
if (mod(year,100) == 0 || mod(year,4) ~= 0) && month == 2 && day <=28
valid = true;
elseif ((mod(year,100) ~= 0 && mod(year,4) == 0) || (mod(year,400) == 0)) && month == 2 && day <= 29
valid = true;
elseif day <= 31 && any(m31(:)== month) == 1
valid = true;
elseif day <= 30 && any(m30(:)== month) == 1
valid = true;
else
valid = false;
end
else
valid = false;
end
end
  1 comentario
Walter Roberson
Walter Roberson el 18 de Jul. de 2021
All those == 1 are unneeded. isscalar() already returns false or true.
The comparisons to 0 are valid, though.

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