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
* )
*/
?>
intdiv
(PHP 7, PHP 8)
intdiv — Целочисленное деление
Описание
intdiv(int
$num1, int $num2): int
Возвращает целочисленное деление num1 на num2.
Список параметров
-
num1 -
Числитель (то, что делится).
-
num2 -
Знаменатель. Числитель будет делиться на
num1
Возвращаемые значения
Целое частное от деления num1 на num2.
Ошибки
Если num2 равен 0, будет вызвано
исключение DivisionByZeroError.
Если num1 равен PHP_INT_MIN,
а num2 равен -1, то будет брошено
исключение ArithmeticError.
Примеры
Пример #1 Пример использования intdiv()
<?php
var_dump(intdiv(3, 2));
var_dump(intdiv(-3, 2));
var_dump(intdiv(3, -2));
var_dump(intdiv(-3, -2));
var_dump(intdiv(PHP_INT_MAX, PHP_INT_MAX));
var_dump(intdiv(PHP_INT_MIN, PHP_INT_MIN));
var_dump(intdiv(PHP_INT_MIN, -1));
var_dump(intdiv(1, 0));
?>
int(1) int(-1) int(-1) int(1) int(1) int(1) Fatal error: Uncaught ArithmeticError: Division of PHP_INT_MIN by -1 is not an integer in %s on line 8 Fatal error: Uncaught DivisionByZeroError: Division by zero in %s on line 9
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User Contributed Notes 6 notes
AmeenRoss ¶
8 years ago
sree dot millu at gmail dot com ¶
4 years ago
@AmeenRoss
This does NOT seem to be equal to intdiv:
<?php
function intdiv_1($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
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.
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)/$b
admin at infis dot net dot ru ¶
4 years ago
For earler versions PHP you may use:
function intdiv_1($a, $b) {
$a = (int) $a;
$b = (int) $b;
return ($a - fmod($a, $b)) / $b;
}
Bubonic dot pestilence at gmail dot com ¶
8 years ago
<?php
function intdiv_2($a, $b) {
$val = $a / $b;
return ($val < 0 ? "ceil" : "floor") ($val);
}
?>
Aren't this?!
