PHP 7.0.6 Released

    intdiv

    (PHP 7)

    intdiv — Integer division

    Description

    int intdiv ( int $dividend , int $divisor )

    Returns the integer quotient of the division of dividend by divisor.

    Parameters

    dividend

    Number to be divided.

    divisor

    Number which divides the dividend.

    Return Values

    The integer quotient of the division of dividend by divisor.

    Errors/Exceptions

    If divisor is 0, a DivisionByZeroError exception is thrown. If the dividend is PHP_INT_MIN and the divisor 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

    User Contributed Notes

    AmeenRoss
    9 months 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
    * )
    */
    ?>
    Bubonic dot pestilence at gmail dot com
    4 months ago
    <?php

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

    ?>

    Aren't this?!
    polettog at gmail dot com
    9 months 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.
    Ts.Saltan
    10 months 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
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