为什么PHP的foreach仅将其数组的指针前进一次(仅)? [英] Why does PHP's foreach advance the pointer of its array (only) once?

问题描述

这是一个好奇问题，关于在PHP中实现foreach的背后原因.

This is a question of curiosity about the reasons behind the way foreach is implemented within PHP.

考虑:

$arr = array(1,2,3);
foreach ($arr as $x) echo current($arr) . PHP_EOL;

它将输出:

2
2
2

我了解到foreach将数组指针倒退到开始处；但是，为什么只增加一次呢?魔术盒子里发生了什么?这只是一件(丑陋的)人工制品吗?

I understand that foreach rewinds array pointers to the beginning; however, why does it then increment it only once? What is happening inside the magic box?? Is this just an (ugly) artefact?

感谢@NickC-对于对zval和refcount感兴趣的其他人，您可以阅读基础知识

Thanks @NickC -- for anyone else curious about zval and refcount, you can read up on the basics here

推荐答案

在第一次迭代之前，$array被软复制"以供在foreach中使用.这意味着没有完成任何实际的复制，但是只有$array的zval的refcount增加到了2.

Right before the first iteration the $array is "soft copied" for use in foreach. This means that no actual copy is done, but only the refcount of the zval of $array is increased to 2.

在第一次迭代中:

1. 该值已提取到$x.
2. 内部数组指针移动到下一个元素，即现在指向2.
3. current通过引用传递的$array进行调用.由于引用的原因，PHP无法再与循环共享zval，因此需要将其分隔(硬拷贝").

1. The value is fetched into $x.
2. The internal array pointer is moved to the next element, i.e. now points to 2.
3. current is called with $array passed by reference. Due to the reference PHP cannot share the zval with the loop anymore and it needs to be separated ("hard copied").

在随后的迭代中，$array zval不再与foreach zval关联.因此，其数组指针不再被修改，并且current始终返回相同的元素.

On the following iterations the $array zval thus isn't anymore related the the foreach zval anymore. Thus its array pointer isn't modified anymore and current always returns the same element.

顺便说一句，我在说明复制行为.在上下文中可能会引起关注，但它与该问题没有直接关系，因为它主要讨论的是硬拷贝.

By the way, I have written a small summary on foreach copying behavior. It might be of interest in the context, but it does not directly relate to the issue as it talks mostly about hard copying.

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