# 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

``` <?phpvar_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 7 notes

39
AmeenRoss
7 years ago
``` This does indeed seem to be equal to intdiv:<?phpfunction intdiv_1(\$a, \$b){    return (\$a - \$a % \$b) / \$b;}?>However, this isn't:<?phpfunction 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 * ) */?> ```
sree dot millu at gmail dot com
3 years ago
``` @AmeenRoss This does NOT  seem to be equal to intdiv:<?phpfunction 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);?>//Output54 ```
oittaa
9 months ago
``` Python style integer division, where the result is always rounded towards minus infinity.1 // 2 is 0(-1) // 2 is -11 // (-2) is -1(-1) // (-2) is 0<?phpfunction 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)); // 0var_dump(intdiv_py(-1, 2)); // -1var_dump(intdiv_py(1, -2)); // -1var_dump(intdiv_py(-1, -2)); // 0?> ```
-21
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) :<?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. ```
-25
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 ```
-11
admin at infis dot net dot ru
3 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;} ```
-36
Bubonic dot pestilence at gmail dot com
6 years ago
``` <?phpfunction intdiv_2(\$a, \$b)  {    \$val = \$a / \$b;    return (\$val < 0 ? "ceil" : "floor") (\$val);} ?>Aren't this?! ```