Difficulties with is_int function in PHP - php

Question

I'm writing a code to solve a problem.one of my functions is not working properly.
function check_factor($sqr,$num)
{
for($i = $sqr ; true ; $i++)
{
$n = pow($sqr,2) - $num;
$s = sqrt($n);
if(is_int($s))
{
return $i;
}
}
}
I know that $s is a "double" , but even when I limit my loop counter to 2,I'll get an endless loop.
What am I missing here? why the function doesnt simply return null? and why I get Infinite loop even when there are 2 iterations?

Answer 1

true is making it an infinite loop
you will always get an infinit loop because sqrt() returns float means
$s = sqrt($n);
$s is a float now
and your test is testing if $s is an integer so it will be an infinit loop even after $i=2 the loop will always stuck and the $i=2 in every loop but
if you change the code to this
<?php
function check_factor($sqr,$num)
{
echo "<br>";
for($i = 1 ; $i < 3 ; $i++)
{
echo " in the loop $i<br>";
$n = pow($sqr,2) - $num;
echo "$n<br>";
$s = sqrt($n);
echo "$s<br>";
if(is_int($s))
{
echo "in the if <br>";
return $i;
}
}
return 0;
}
$val=0;
$val = check_factor(5,2);
echo "<br>$val<br>";
?>
the out put should be like this
in the loop 1
23
4.7958315233127
in the loop 2
23
4.7958315233127
0
and that's it. i hope i helped.

Answer 2

According to the manual: PHP function sqrt() always returns a float and NEVER an integer.
Check http://php.net/manual/en/function.sqrt.php