[ACCEPTED]-How do I test for integers in MATLAB?-integer

You can use the mod function, which returns 6 the remainder after division. All integers 5 are divisible by 1. So a good test for non-integer 4 would be

integerTest=~mod(value,1);

This returns 0 if value is not an integer 3 and 1 if it is. You can then use this as 2 a conditional to reject non-integer user 1 inputs.

Here is another variation (you can see it 2 being used in ISIND function: edit isind.m):

integerTest = ( x == floor(x) );

On my machine, it 1 is faster than the other proposed solutions:

%# create a vector of doubles, containing integers and non-integers
x = (1:100000)';                       %'
idx = ( rand(size(x)) < 0.5 );
x(idx) = x(idx) + rand(sum(idx),1);

%# test for integers
tic, q1 = ~mod(x, 1); toc
tic, q2 = x==double(uint64(x)); toc
tic, q3 = x==floor(x); toc

%# compare results
assert( isequal(q1,q2,q3) )

Timings:

Elapsed time is 0.012253 seconds.
Elapsed time is 0.014201 seconds.
Elapsed time is 0.005665 seconds.

You can cast the value to an integer and 4 back to a double and check the result against 3 the original value:

>> x = 1.3;
>> x == double(uint64(x))

ans =

     0

>> x = 2;
>> x == double(uint64(x))

ans =

     1

Interestingly, R.M.'s approach of using 2 MOD runs faster in a loop and the above 1 casting approach runs faster when vectorized:

>> x = rand(100000, 1);
>> tic; for ii = 1:100000; ~mod(x(ii), 1); end; toc;
Elapsed time is 0.018380 seconds.
>> tic; for ii = 1:100000; x(ii) == double(uint64(x(ii))); end; toc;
Elapsed time is 0.383020 seconds.
>> tic; ~mod(x, 1); toc;
Elapsed time is 0.005299 seconds.
>> tic; x == double(uint64(x)); toc;
Elapsed time is 0.002971 seconds.

assert(isnumeric(input) && round(input) == input, 'That number is not an integer.')

You could add other checks, (like for positivity) easily 4 as well.

Edited using isinteger. Thanks @SolarStatistics, I hadn't 3 noticed they added this functionality. Edited 2 back to original answer again as isinteger isn't 1 appropriate (see comments below).

As point out by @nibot isinteger tests for the input 3 as an integer TYPE. Instead you could check 2 to see if rounding input returns the same value 1 as input. eg:

assert(abs(round(input)-input))<eps*2,'That number is not an integer.')

for example

>> input=1.3;
>> assert(abs(round(input)-input)<eps*2,'That number is not an integer.')
??? That number is not an integer.

>> input=3;
>> assert(abs(round(input)-input)<eps*2,'That number is not an integer.')
>> 

I just wanted to point out that the provided 12 methods all test for whether the input is 11 a Gaussian integer, meaning that the real and imaginary 10 parts are both integers. If you need to care 9 about the imaginary part then you need to 8 deal with it separately.

For my applications, inputs 7 with imaginary components shouldn't be considered 6 a valid integer, so I have this:

function boolResult = fnIsInteger(input)
    %validate input
    if isempty(input)
        error('Input cannot be empty')
    elseif ~isnumeric(input)
        error('Input must be numeric')
    end

    boolResult = (imag(input) == 0) & (round(input) == input);
end

Using b3.'s 5 tests:

>> x = rand(100000, 1);
>> tic; for ii = 1:100000; ~mod(x(ii), 1); end; toc;
Elapsed time is 0.003960 seconds.
>> tic; for ii = 1:100000; fnIsInteger(x(ii)); end; toc;
Elapsed time is 0.217397 seconds.
>> tic; ~mod(x, 1); toc;
Elapsed time is 0.000967 seconds.
>> tic; fnIsInteger(x); toc;
Elapsed time is 0.003195 seconds.

The looped call is quite a bit slower 4 mostly due to the function overhead. Replacing 3 the arithmetic expression with ~mod(dataInput, 1) will 2 make it only 50% faster than the code that 1 checks for imaginary parts.