foreach循环和与放大器基准; $值 [英] foreach loop and reference of &$value

问题描述

为什么空foreach循环可以改变结果。

Why is an empty foreach loop can change the result.

我有以下的code：

  $variable = [1,2,3,4];
  foreach ($variable  as $key => &$value) 
    $value ++;
  var_dump($variable);

结果我得到的是

  array (size=4)
  0 => int 2
  1 => int 3
  2 => int 4
  3 => &int 5

现在，当我添加一个空的foreach循环像这样

Now ,when I add an empty foreach loop like this

  $variable  = [1,2,3,4];
  foreach ($variable  as $key => &$value) 
    $value ++;
  foreach ($variable  as $key => $value);
  var_dump($variable);

我得到这样的：

 array (size=4)
 0 => int 2
 1 => int 3
 2 => int 4
 3 => &int 4

有人可以解释我为什么当我添加第二个空循环，最后一个元素不会改变为什么有急症室;最后一个元素的盈？

can someone explain me why the last element doesn't change when I add the second empty loop, and why there is a & infront of the last element ?

推荐答案

在第一循环结束， $值指向同一个地方 $变量[3] （它们都指向同一个内存位置）

At the end of the first loop, $value is pointing to the same place as $variable[3] (they are pointing to the same location in memory):

$variable  = [1,2,3,4];
foreach ($variable  as $key => &$value) 
    $value ++;

即使在这个循环结束后， $值仍是的指向内存中的相同位置 $引用变量[3] ，让您保存在 $值的值，每次，这也将覆盖存储 $变量的值[3] ：

Even as this loop is finished, $value is still a reference that's pointing to the same location in memory as $variable[3], so each time you store a value in $value, this also overwrites the value stored for $variable[3]:

foreach ($variable as $key => $value);
var_dump($variable);

通过本的foreach的每一个评价，既 $值和 $变量[3] 正在成为等于在$变量可迭代项目的价值。

With each evaluation of this foreach, both $value and $variable[3] are becoming equal to the value of the iterable item in $variable.

因此​​，在第二循环的第三次迭代， $值和 $变量[3] 相等4参照，那么第四和第二循环的最后一次迭代过程中，没有什么变化，因为你传递的 $变量[3] （这仍然是<$值C $ C>＆安培; $值）以 $值（这仍然是＆安培; $值）。

So in the 3rd iteration of the second loop, $value and $variable[3] become equal to 4 by reference, then during the 4th and final iteration of the second loop, nothing changes because you're passing the value of $variable[3] (which is still &$value) to $value (which is still &$value).

这是非常令人困惑，但它并不甚至略有特质;它是code执行的究竟的，因为它应该。

It's very confusing, but it's not even slightly idiosyncratic; it's the code executing exactly as it should.

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