# intdiv

(PHP 7, PHP 8)

intdivDivisión entera

### Descripción

intdiv(int \$dividend, int \$divisor): int

Devuelve el cociente entero de la división de dividend entre divisor.

### Parámetros

dividend

El número a dividir.

divisor

El número que divide a dividend

### Valores devueltos

El cociente entero de la división de dividend entre divisor.

### Errores/Excepciones

Si divisor es 0, se lanza una excepción DivisionByZeroError. Si dividend es PHP_INT_MIN y divisor es -1, entonces se lanza una excepción ArithmeticError.

### Ejemplos

Ejemplo #1 Ejemplo de intdiv()

<?php
var_dump
(intdiv(32));
var_dump(intdiv(-32));
var_dump(intdiv(3, -2));
var_dump(intdiv(-3, -2));
var_dump(intdiv(PHP_INT_MAXPHP_INT_MAX));
var_dump(intdiv(PHP_INT_MINPHP_INT_MIN));
var_dump(intdiv(PHP_INT_MIN, -1));
var_dump(intdiv(10));
?>
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 6 notes

39
AmeenRoss
6 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
* )
*/
?>
sree dot millu at gmail dot com
2 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
-19
polettog at gmail dot com
6 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.
-26
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
-7
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;
}
-33
Bubonic dot pestilence at gmail dot com
6 years ago
<?php

function intdiv_2(\$a, \$b)  {

\$val = \$a / \$b;
return (
\$val < 0 ? "ceil" : "floor") (\$val);
}

?>

Aren't this?!