sum inside if statement in PHP

Question

somebody can explain me why this code:

$a = 0.1;
$b = 0.2;

if ($a + $b == 0.3) {
    echo "OK";
} else {
    echo "KO";
}

returns KO?

I don't understand why the sum result is different the float 0.3, considering that:

var_dump($a + $b);

returns: float(0.3)

The only hypothesis I have is that the comparison is made between only $b and 0.3 but the doubt remains because also in this case:

if ( ($a + $b) == 0.3) {

I get KO..

If you set the precision than the comparison will be done, otherwise not. — Murad Hasan, May 30 '16 at 09:00
See this example: [https://3v4l.org/0Ad6A](https://3v4l.org/0Ad6A) — Murad Hasan, May 30 '16 at 09:03
https://en.wikipedia.org/wiki/Floating_point#Accuracy_problems — Federkun, May 30 '16 at 09:08
Offtopic: how do you not get confused when using `OK` and `KO`. I mean, I would often overlook the difference. — Peon, May 30 '16 at 09:09
[http://php.net/manual/en/language.types.float.php#113703](http://php.net/manual/en/language.types.float.php#113703) answers perfectly to the question. Thanks — suikoy, May 30 '16 at 09:13

score 0 · Answer 1 · answered May 30 '16 at 09:04

This is a floating point number and the precision of this number is not fixed, if you fixed the precision than it will be okey.

$a = 0.1;
$b = 0.2;
$c = sprintf('%.1f', ($a + $b));
if ($c  == 0.3) {
    echo "OK";
} else {
    echo "KO";
}

the result is OK now.

score 0 · Answer 2 · answered May 30 '16 at 09:10

0

Use function round($a+b, 1); I have this problem in my code

answered May 30 '16 at 09:10

Pudich Denis

2 Answers2