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)
intdiv — Integer division
Descrierea
intdiv
( int
$num1
, int $num2
) : int
Returns the integer quotient of the division of num1 by num2.
Parametri
-
num1 -
Number to be divided.
-
num2 -
Number which divides the
num1.
Valorile întoarse
The integer quotient of the division of num1 by num2.
Erori/Excepții
If num2 is 0, a DivisionByZeroError
exception is thrown. If the num1 is PHP_INT_MIN
and the num2 is -1, then an
ArithmeticError exception is thrown.
Exemple
Example #1 intdiv() example
<?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 7 notes
AmeenRoss ¶
9 years ago
oittaa ¶
2 years 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, $num2);
}
var_dump(intdiv_py(1, 2)); // 0
var_dump(intdiv_py(-1, 2)); // -1
var_dump(intdiv_py(1, -2)); // -1
var_dump(intdiv_py(-1, -2)); // 0
?>
sree dot millu at gmail dot com ¶
5 years ago
@AmeenRoss
This does NOT seem to be equal to intdiv:
<?php
function intdiv_1($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 ¶
9 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 $b really needs to be of type integer though.
If they are too big and indivisible, some precision will be lost during the conversion to float and the result may be inaccurate.
admin at infis dot net dot ru ¶
5 years ago
For earler versions PHP you may use:
function intdiv_1($a, $b) {
$a = (int) $a;
$b = (int) $b;
return ($a - fmod($a, $b)) / $b;
}
Ts.Saltan ¶
9 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)/$b
Bubonic dot pestilence at gmail dot com ¶
8 years ago
<?php
function intdiv_2($a, $b) {
$val = $a / $b;
return ($val < 0 ? "ceil" : "floor") ($val);
}
?>
Aren't this?!
