How do you divide in php?

(PHP 7, PHP 8)

intdiv — Integer division

Description

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

Parameters

Number to be divided.

Number which divides the num1.

Return Values

The integer quotient of the division of num1 by num2.

Examples

Example #1 intdiv() example

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() - Floating-point modulo

AmeenRoss ¶

7 years ago

This does indeed seem to be equal to intdiv:

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

However, this isn't:

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

Consider an example where either of the parameters is negative:
$param1 = -10;
$param2 = 3;
print_r([
    'modulus' => intdiv_1($param1, $param2),
    'floor' => intdiv_2($param1, $param2),
]);/**
* Array
* (
*     [modulus] => -3
*     [floor] => -4
* )
*/
?>

sree dot millu at gmail dot com ¶

2 years ago

@AmeenRoss
This does NOT  seem to be equal to intdiv:

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

See this example code


$x

oittaa ¶

7 months 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

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
?>

polettog at gmail dot com ¶

7 years ago

Without intdiv(), the following may be a good way (with $a and $b of type integer and not too big) :
(int)($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.

Ts.Saltan ¶

7 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

admin at infis dot net dot ru ¶

2 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;
}