(PHP 7, PHP 8)
intdiv — 对除法结果取整
$num1, int $num2): int
返回 num1 除以 num2 商数的整数部分。
num1被除数。
num2除数。
num1 除以 num2 的商，对该商取整。
如果 num2 是 0，将抛出
DivisionByZeroError 异常。如果 num1
是 PHP_INT_MIN 并且 num2 是
-1，将抛出 ArithmeticError 异常.
示例 #1 intdiv() 的一些示例
<?php
var_dump(intdiv(3, 2));
var_dump(intdiv(-3, 2));
var_dump(intdiv(3, -2));
var_dump(intdiv(-3, -2));
var_dump(intdiv(PHP_INT_MAX, PHP_INT_MAX));
var_dump(intdiv(PHP_INT_MIN, PHP_INT_MIN));
var_dump(intdiv(PHP_INT_MIN, -1));
var_dump(intdiv(1, 0));
?>int(1) int(-1) int(-1) int(1) int(1) int(1) Fatal error: Uncaught ArithmeticError: Division of PHP_INT_MIN by -1 is not an integer in %s on line 8 Fatal error: Uncaught DivisionByZeroError: Division by zero in %s on line 9
This does indeed seem to be equal to intdiv:
<?php
function intdiv_1($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, $num2);
}
var_dump(intdiv_py(1, 2)); // 0
var_dump(intdiv_py(-1, 2)); // -1
var_dump(intdiv_py(1, -2)); // -1
var_dump(intdiv_py(-1, -2)); // 0
?>@AmeenRoss
This does NOT seem to be equal to intdiv:
<?php
function intdiv_1($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
4Without 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 $b really needs to be of type integer though.
If they are too big and indivisible, some precision will be lost during the conversion to float and the result may be inaccurate.For earler versions PHP you may use:
function intdiv_1($a, $b) {
$a = (int) $a;
$b = (int) $b;
return ($a - fmod($a, $b)) / $b;
}$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)/$b<?php
function intdiv_2($a, $b) {
$val = $a / $b;
return ($val < 0 ? "ceil" : "floor") ($val);
}
?>
Aren't this?!