(PHP 7, PHP 8)
intdiv — 对除法结果取整
num1被除数。
num2除数。
num1 除以 num2 的商,对该商取整。
如果 num2 是 0,将抛出
DivisionByZeroError 异常。如果 num1
是 PHP_INT_MIN 并且 num2 是
-1,将抛出 ArithmeticError 异常.
示例 #1 intdiv() 的一些示例
<?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));
?>以上示例会输出:
int(1) int(-1) int(-1) int(1) int(1) int(1)
示例 #2 intdiv() 无效除数的示例
<?php
try {
intdiv(PHP_INT_MIN, -1);
} catch (Error $e) {
echo get_class($e), ': ', $e->getMessage(), PHP_EOL;
}
try {
intdiv(1, 0);
} catch (Error $e) {
echo get_class($e), ': ', $e->getMessage(), PHP_EOL;
}
?>以上示例会输出:
ArithmeticError: Division of PHP_INT_MIN by -1 is not an integer DivisionByZeroError: Division by zero
This does indeed seem to be equal to intdiv:
<?php
function intdiv_1($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
* )
*/
?>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
?>