Understanding something more about the % Modulus operator

Question

I am learning to work with some math like PHP query and just got to the modulo, I am not quite sure in what situations to use this because of something i stumbled on and yes I did already read one of the posts here about the modulo : Understanding The Modulus Operator %

(This explanation is only for positive numbers since it depends on the language otherwise)

The quote above is in the top answer there. But if I focus on PHP only and i use the modulo like this:

$x = 8;
$y = 10;
$z = $x % $y;
echo $z; // this outputs 8 and I semi know why.

Calculation: (8/10) 0 //times does 10 fit in 8.
                    0 * 10 = 0 //So this is the number that has to be taken off of the 8
                    8 - 0 = 8 //<-- answer

Calculation 2: (3.2/2.4) 1 //times does this fit
                         1 * 2.4 = 2.4 //So this is the number that has to be taken off of the 3.2
                         3.2 - 2.4 = 0.8 // but returns 1?

So my question is why does this exactly happen. my guess would be that in the first phase it would get 8/10 = 0,8 but this doesn't happen. So can someone explain a bit about why this happens. I understand the modulo's basics like if I do 10 % 8 = 2 and I semi understand why it doesn't return something like this: 8 % 10 = -2.

Also, is there a way to modify how the modulo works? so it would return a - value or a decimal value in the calculation? or would I need to use something else for this

Little shortened: why does this happen when I get a negative number in return and is there some other way or operator that can actually do the same and get in the negative numbers.

From the manual % - 'Remainder of $a divided by $b.', so 8/10 = 0 (only integer part), remainder = 8. — Nigel Ren, Feb 07 '18 at 11:07
@NigelRen I did read it but is there a way for it to actually go to something like a double? or to the negative. becouse this isnt talked about in the manual. (as far as ive seen) — B. Dionys, Feb 07 '18 at 11:08
As for -ve - from manual page `echo (-5 % 3)."\n"; // prints -2` — Nigel Ren, Feb 07 '18 at 11:10
For floating point modulus, there is a function [fmod()](http://php.net/manual/en/function.fmod.php) — Mark Baker, Feb 07 '18 at 11:14
@NigelRen hmm oke I hadnt thought about that yet but its still int then right? — B. Dionys, Feb 07 '18 at 11:22
@MarkBaker Ive read the page and yes idk how I didnt find this yet but this is somewhat wat I thought should have been possible with another function — B. Dionys, Feb 07 '18 at 11:23

score 2 · Accepted Answer · answered Feb 07 '18 at 11:15

2

Modulus (%) only works for integers, so your calculation at the bottom of your example is correct...

8/10 = 0 ( integer only ), remainder = 8-(0*10) = 8.

If you instead had -ve 12 - -12%10...

-12/10 = -1 (again integer only), remainder = -12 - (10*-1) = -2

For floats - you can use fmod(http://php.net/manual/en/function.fmod.php)

<?php
$x = 5.7;
$y = 1.3;
$r = fmod($x, $y);
// $r equals 0.5, because 4 * 1.3 + 0.5 = 5.7

(Example from manual)

answered Feb 07 '18 at 11:15

Nigel Ren

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Ah oke, thanks for this like @markbaker also said in the comments, this is what I thought should be doable but I hadnt found this function yet. maybe becouse I searched wrong or something but thanks this explains and helps a lot. – B. Dionys Feb 07 '18 at 11:25

Understanding something more about the % Modulus operator

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