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intdiv

(PHP 7, PHP 8)

intdivInteger division

Description

intdiv(int $num1, int $num2): int

Returns the integer quotient of the division of num1 by num2.

Parameters

num1

Number to be divided.

num2

Number which divides the num1.

Return Values

The integer quotient of the division of num1 by num2.

Errors/Exceptions

If num2 is 0, a DivisionByZeroError exception is thrown. If the num1 is PHP_INT_MIN and the num2 is -1, then an ArithmeticError exception is thrown.

Examples

Example #1 intdiv() example

<?php
var_dump
(intdiv(3, 2));
var_dump(intdiv(-3, 2));
var_dump(intdiv(3, -2));
var_dump(intdiv(-3, -2));
var_dump(intdiv(PHP_INT_MAX, PHP_INT_MAX));
var_dump(intdiv(PHP_INT_MIN, PHP_INT_MIN));
var_dump(intdiv(PHP_INT_MIN, -1));
var_dump(intdiv(1, 0));
?>
int(1)
int(-1)
int(-1)
int(1)
int(1)
int(1)

Fatal error: Uncaught ArithmeticError: Division of PHP_INT_MIN by -1 is not an integer in %s on line 8
Fatal error: Uncaught DivisionByZeroError: Division by zero in %s on line 9

See Also

  • / - Floating-point division
  • % - Integer modulus
  • fmod() - Returns the floating point remainder (modulo) of the division of the arguments - Floating-point modulo

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User Contributed Notes 7 notes

up
44
AmeenRoss
8 years ago
This does indeed seem to be equal to intdiv:

<?php
function intdiv_1($a, $b){
return (
$a - $a % $b) / $b;
}
?>

However, this isn't:

<?php
function intdiv_2($a, $b){
return
floor($a / $b);
}
?>

Consider an example where either of the parameters is negative:
<?php
$param1
= -10;
$param2 = 3;
print_r([
'modulus' => intdiv_1($param1, $param2),
'floor' => intdiv_2($param1, $param2),
]);

/**
* Array
* (
* [modulus] => -3
* [floor] => -4
* )
*/
?>
up
6
oittaa
2 years ago
Python style integer division, where the result is always rounded towards minus infinity.

1 // 2 is 0
(-1) // 2 is -1
1 // (-2) is -1
(-1) // (-2) is 0

<?php
function intdiv_py(int $num1, int $num2): int{
if (
$num1 < 0 xor $num2 < 0){
$num1 = abs($num1);
$num2 = abs($num2);
$remainder = $num1 % $num2;
return
$remainder ? -1 -($num1 - $remainder) / $num2 : -$num1 / $num2;
}
return
intdiv($num1, $num2);
}

var_dump(intdiv_py(1, 2)); // 0
var_dump(intdiv_py(-1, 2)); // -1
var_dump(intdiv_py(1, -2)); // -1
var_dump(intdiv_py(-1, -2)); // 0
?>
up
-1
sree dot millu at gmail dot com
4 years ago
@AmeenRoss
This does NOT seem to be equal to intdiv:

<?php
function intdiv_1($a, $b){
return (
$a - $a % $b) / $b;
}
?>

See this example code
<?php

$x
= 5.6;
$y = 1.4 ;

echo
intdiv($x,$y);

echo
"\n";

function
intdiv_1($a, $b){
return (
$a - $a % $b) / $b;
}

echo
intdiv_1($x,$y);
?>

//Output
5
4
up
-17
polettog at gmail dot com
8 years ago
Without intdiv(), the following may be a good way (with $a and $b of type integer and not too big) :
<?php
(int)($a / $b)
?>
because in case of divisible integers, the result will be integer and there is no risk of float appearing round but below their represented value (like the case (0.1+0.7)*10).
$a and $b really needs to be of type integer though.
If they are too big and indivisible, some precision will be lost during the conversion to float and the result may be inaccurate.
up
-8
admin at infis dot net dot ru
4 years ago
For earler versions PHP you may use:

function intdiv_1($a, $b) {
$a = (int) $a;
$b = (int) $b;
return ($a - fmod($a, $b)) / $b;
}
up
-24
Ts.Saltan
8 years ago
$a = 57;
$b = 3;

var_dump(
intdiv($a,$b),
intdiv_1($a,$b),
intdiv_2($a,$b)
);

function intdiv_1($a, $b){
return ($a-$a%$b)/$b;
}

function intdiv_2($a, $b){
return floor($a/$b);
}
//intdiv($a, $b) == floor($a/$b) == ($a-$a%$b)/$b
up
-36
Bubonic dot pestilence at gmail dot com
8 years ago
<?php

function intdiv_2($a, $b) {
$val = $a / $b;
return (
$val < 0 ? "ceil" : "floor") ($val);
}

?>

Aren't this?!
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