Why does PHP's foreach advance the pointer of its array (only) once?

Question

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

Consider:

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

which will output:

2
2
2

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?

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

Answer 1

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.

On the first iteration:

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").

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.

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.