intdiv()

版本：php7

(PHP 7)

对除法结果取整

说明

返回$dividend除以$divisor商数的整数部分。

参数

被除数。

除数。

返回值

$dividend除以$divisor的商，对该商取整。

错误／异常

如果$divisor是0，将抛出DivisionByZeroError异常。如果$dividend是PHP_INT_MIN并且$divisor是-1，将抛出ArithmeticError异常.

范例

Example #1intdiv()的一些例子

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:

However, this isn't:

Consider an example where either of the parameters is negative:

@AmeenRoss 
This does NOT seem to be equal to intdiv:

See this example code

//Output
5
4

For earler versions PHP you may use:
function intdiv_1($a, $b) {
  $a = (int) $a;
  $b = (int) $b;
  return ($a - fmod($a, $b)) / $b;
}

Without intdiv(), the following may be a good way (with $a and $b of type integer and not too big) :

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.

$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

Aren't this?!