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php源码调试foreach诡异现象

Jun 06, 2016 pm 08:33 PM
php

$array = array(1,2,3,4);
foreach($array as $key => $value) {
$value = &$array[$key];
echo current($array);
}
输出 2 2 3 3
这段代码给我搞懵了,希望大神通过php源码解释一下。

回复内容:

$array = array(1,2,3,4);
foreach($array as $key => $value) {
$value = &$array[$key];
echo current($array);
}
输出 2 2 3 3
这段代码给我搞懵了,希望大神通过php源码解释一下。

首先,要理解这个问题,我们得先将循环拆解,然后看看数组的值会怎么变化:

1) 第一次循环

<code>php</code><code>$value = $array[0];
$value = &$array[0]; // $value现在变成了$array[0]的引用了
</code>
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这时的$array = array(1, 2, 3, 4);

2) 第二次循环

<code>php</code><code>$value = $array[1]; // 由于$value引用$array[0],此语句相当于$array[0] = $array[1],也就是$array[0] = 2
$value = &$array[1]; // $value现在变成了$array[1]的引用了
</code>
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这时的$array = array(2, 2, 3, 4);

3) 第三次循环

<code>php</code><code>$value = $array[2]; // 由于$value引用$array[1],此语句相当于$array[1] = $array[2],也就是$array[1] = 3
$value = &$array[2]; // $value现在变成了$array[2]的引用了
</code>
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这时的$array = array(2, 3, 3, 4);

4) 第四次循环

<code>php</code><code>$value = $array[3]; // 由于$value引用$array[2],此语句相当于$array[2] = $array[3],也就是$array[2] = 4
$value = &$array[3]; // $value现在变成了$array[3]的引用了
</code>
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这时的$array = array(2, 3, 4, 4);

现在我们来看一下脚本为什么会输出2233,题主可以试一下在foreach内部使用key($array)来看看它的返回值是什么,你会发现key()函数返回的值都是1,也就是说,数组的指针永远都是指向下标为1的元素,所以在每次循环中,current($array)函数的返回值也永远是下标为1的元素的值。

最后,我们来看看每次循环中,数组下标为1的元素的值是什么?是的,就是下面图中的这些:
php源码调试foreach诡异现象

至于,为什么key()函数的返回值永远都是1呢,题主可以参考这个网址中的“问题2”,里面有详细的解释:)

其实也搞不懂,因为好奇,就研究了下。

$array = array(1,2,3,4);
foreach($array as $key => $value) {
echo current($array);
}

看看这个结果,也许你会诧异,我也诧异,指针并没有变化。答案是:2 2 2 2

所以current($array)永远只会输出这个数组的下标为1的值。之所以输出2 2 3 3 ,你现在也知道,$value = &$array[$key];这句改变了原来数组的值了。

$array = array(1,2,3,4);
foreach($array as $key => $value) {
$value = &$array[$key];
echo current($array);
var_dump($array);
}

再输出看看,果然然来的数组的值一直在改变。问题又来了,那么为什么会变呢?

http://segmentfault.com/q/1010000002676308
参考这个,这个问题懂了,你就自然懂了。

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