intdiv

(PHP 7, PHP 8)

intdivЦелочисленное деление

Описание

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

Возвращает результат целочисленного деления num1 на num2.

Список параметров

num1

Числитель (то, что делится).

num2

Знаменатель. Число, на которое делится num1

Возвращаемые значения

Целое частное от деления num1 на num2.

Ошибки

Если num2 равен 0, будет выброшено исключение DivisionByZeroError. Если num1 равен PHP_INT_MIN, а num2 равен -1, то будет выброшено исключение ArithmeticError.

Примеры

Пример #1 Пример использования intdiv()

<?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

Смотрите также

  • / - деление рациональных чисел
  • % - остаток от деления
  • fmod() - остаток от деления рациональных чисел

add a note

User Contributed Notes 7 notes

up
41
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
1
oittaa
1 year 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
0
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
-20
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
-9
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
-25
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
7 years ago
<?php

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

?>

Aren't this?!
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