Why is the output of sqrt
not an integer for "16" in PHP?
Example
php > $fig = 16;
php > $sq = sqrt($fig); //should be 4
php > echo $sq;
4
php > echo is_int($sq); // should give 1, but gives false
php >
I feel that the problem is in the internal presentation which PHP hides similarly as Python. How can you then know when the given figure is integer after taking a square root?
So how can you differentiate between 4
and 4.12323
in PHP without using a regex?