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php – Why would is_int(sqrt(100)) return false

Posted by: admin July 12, 2020 Leave a comment


Does anyone know why this always returns False?


and what syntax should i use to check is square root is an integer ?

How to&Answers:

If you want to check whether a sqrt is an integer, you can do:

$root = sqrt($val);

if((int) $root == $root)) {
    // root is integer


sqrt function returns float value, not int.


The problem here is in your definition of "integer". You’re reading it as "a value with no non-zero fractional significant figures", whereas in is_int it just refers to a datatype, i.e. literally anything with type int.

That is, the floating-point value 10.0 is still a floating-point value, even though its mathematical value is equal to that of the integer 10.

The result of sqrt is never an integer; however, you can check whether a floating-point value has any non-zero fractional significant figures by comparing it to one in which you deliberately took them all off.

A naive implementation:

$sqrt = sqrt(100);
if ($sqrt == (int)$sqrt) {
   // ...

However, of course you should never compare anything against a floating-point value with ==; use your favourite floating-point equality mechanism to perform this test.


Actually, I suppose the only values for which this comparison could potentially fail are fractional, in which case the test should fail anyway. So == might be sufficient.


sqrt returns a float, not an integer, therefore is_int returns false.

Try use is_float instead.


is_int() refers to the datatype of the result, not to the value contained in that result. sqrt() always returns a result of datatype float, irrespective of the value


Data type of sqrt(100) is float. is_int checks for the data type not the value