PHP : 2258574.18 not equals to 2058555.18 + 200019 - 0

Question

I'm struggling to make accounting software in PHP. And i got some bug from float precision.

Here is the sample unworking code :

$a = (float) 2258574.18;
$b = (float) 2058555.18;
$c = 200019;
$d = 0;

($b+$c-$d == $a ) ? $x = "equals" : $x = "!equals"; 
echo $x;

Outputs :

!equals

I was using round() abs() but doesnt solved. It work if i only convert it to int (int).

I almost suicide because of this.

do `echo "a:$a b:$b";` and you will see how these values are stored internally. — Peter VARGA, Jun 02 '16 at 06:58
FYI: In console (js) doing `2058555.18 + 200019` will output `2258574.1799999997` — Justinas, Jun 02 '16 at 07:10
You do not want to write an accounting system using floats! Use a library which supports decimal numbers, or do all calculations in ints. — deceze, Jun 02 '16 at 07:40

score 0 · Accepted Answer · answered Jun 02 '16 at 07:05

0

Try round() with precision 2

$e = $b+$c-$d;   
(round($e, 2) == round($a, 2) ) ? $x = "equals" : $x = "!equals"; 
echo $x;

answered Jun 02 '16 at 07:05

ASR

Is it good to round accounting value ? I think not. – code K.O. Jun 02 '16 at 07:18

1 Answers1