What are the significant differences with these two PHP equations?

Question

You have two variables namely $a = 5 and $b = 9

What are the major mathematical differences between these two statements?

floor(floor($a / $b) - .5); //output: -1

and

(int)((int)($a / $b) - .5); //output: 0

Guess: `floor` rounds down (to minus infinity), `(int)` truncates (rounds toward 0). — zch, Dec 21 '13 at 20:06
The tope answer to [this question](http://stackoverflow.com/questions/3300290/cast-to-int-vs-floor) explains that. Same as zch's comment above. — Michael, Dec 21 '13 at 20:19

score 3 · Accepted Answer · answered Dec 21 '13 at 20:17

3

Look to the rounding of parts of your expression.

The division of 5/9 is

$a / $b = 5 / 9 = 0.555555556

floor(1.5) = 1
floor(-1.5) = -2

then

floor(floor($a / $b) - .5) = floor(floor(0.555555556)) = floor(0 - .5) = floor(-.5) = -1

(int) 4.32 = 4

then

(int)((int)($a / $b) - .5) = (int)((int)(0.555555556) - .5) = (int)(0 - .5) = int(-0.5) = 0;

answered Dec 21 '13 at 20:17

Martin Strejc

That's wrong. That's not intval, that's typecasting: look here!: http://php.net/manual/de/language.types.type-juggling.php – Sebi2020 Dec 21 '13 at 22:09
@Sebi2020 Right, even though that's not 'intval', the rounding algorithm is the same. I used the link because there is the explanation of rounding on php.net. If you've a better one edit the answer please. – Martin Strejc Dec 21 '13 at 22:17
Apparently I'm not. I got two downvotes for my answer. – Sebi2020 Dec 21 '13 at 22:23

Sebi2020 · Answer 2 · 2013-12-21T22:09:08.700

-2

(int) is a typecast and you get an Integer e.g. 2, 3 , 5 etc. floor rounds down.

So with (int)((int) 5 / 9)-0.5) you get:

0 - 0 (0.56666666) = 0

(int) isn't a function! Look here: Type Casting - PHP DOC

edited Dec 21 '13 at 22:09

answered Dec 21 '13 at 20:10

Sebi2020

2 Answers2