Home How does PHP 'foreach' actually work?
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How does PHP 'foreach' actually work?

user14573
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user14573 Published in September 19, 2018, 6:44 pm

Let me prefix this by saying that I know what foreach is, does and how to use it. This question concerns how it works under the bonnet, and I don't want any answers along the lines of "this is how you loop an array with foreach".


For a long time I assumed that foreach worked with the array itself. Then I found many references to the fact that it works with a copy of the array, and I have since assumed this to be the end of the story. But I recently got into a discussion on the matter, and after a little experimentation found that this was not in fact 100% true.

Let me show what I mean. For the following test cases, we will be working with the following array:

$array = array(1, 2, 3, 4, 5);

Test case 1:

foreach ($array as $item) {
  echo "$item\n";
  $array[] = $item;
}
print_r($array);

/* Output in loop:    1 2 3 4 5
   $array after loop: 1 2 3 4 5 1 2 3 4 5 */

This clearly shows that we are not working directly with the source array - otherwise the loop would continue forever, since we are constantly pushing items onto the array during the loop. But just to be sure this is the case:

Test case 2:

foreach ($array as $key => $item) {
  $array[$key + 1] = $item + 2;
  echo "$item\n";
}

print_r($array);

/* Output in loop:    1 2 3 4 5
   $array after loop: 1 3 4 5 6 7 */

This backs up our initial conclusion, we are working with a copy of the source array during the loop, otherwise we would see the modified values during the loop. But...

If we look in the manual, we find this statement:

When foreach first starts executing, the internal array pointer is automatically reset to the first element of the array.

Right... this seems to suggest that foreach relies on the array pointer of the source array. But we've just proved that we're not working with the source array, right? Well, not entirely.

Test case 3:

// Move the array pointer on one to make sure it doesn't affect the loop
var_dump(each($array));

foreach ($array as $item) {
  echo "$item\n";
}

var_dump(each($array));

/* Output
  array(4) {
    [1]=>
    int(1)
    ["value"]=>
    int(1)
    [0]=>
    int(0)
    ["key"]=>
    int(0)
  }
  1
  2
  3
  4
  5
  bool(false)
*/

So, despite the fact that we are not working directly with the source array, we are working directly with the source array pointer - the fact that the pointer is at the end of the array at the end of the loop shows this. Except this can't be true - if it was, then test case 1 would loop forever.

The PHP manual also states:

As foreach relies on the internal array pointer changing it within the loop may lead to unexpected behavior.

Well, let's find out what that "unexpected behavior" is (technically, any behavior is unexpected since I no longer know what to expect).

Test case 4:

foreach ($array as $key => $item) {
  echo "$item\n";
  each($array);
}

/* Output: 1 2 3 4 5 */

Test case 5:

foreach ($array as $key => $item) {
  echo "$item\n";
  reset($array);
}

/* Output: 1 2 3 4 5 */

...nothing that unexpected there, in fact it seems to support the "copy of source" theory.


The Question

What is going on here? My C-fu is not good enough for me to able to extract a proper conclusion simply by looking at the PHP source code, I would appreciate it if someone could translate it into English for me.

It seems to me that foreach works with a copy of the array, but sets the array pointer of the source array to the end of the array after the loop.

  • Is this correct and the whole story?
  • If not, what is it really doing?
  • Is there any situation where using functions that adjust the array pointer (each(), reset() et al.) during a foreach could affect the outcome of the loop?
share|improve this question
  • 3
    @DaveRandom There's a php-internals tag this should probably go with, but I'll leave it to you to decide which if any of the other 5 tags to replace. – Michael Berkowski Apr 7 '12 at 19:40
  • 4
    looks like COW, without delete handle – zb' Apr 7 '12 at 19:43
  • 114
    At first I thought »gosh, another newbie question. Read the docs… hm, clearly undefined behavior«. Then I read the complete question, and I must say: I like it. You've put quite some effort in it and writing all the testcases. ps. are testcase 4 and 5 the same? – knittl Apr 7 '12 at 19:49
  • 19
    Just a thought about why it does make sense that the array pointer gets touched: PHP needs to reset and move the internal array pointer of the original array along with the copy, because the user may ask for a reference to the current value (foreach ($array as &$value)) - PHP needs to know the current position in the original array even though it's actually iterating over a copy. – Niko Apr 7 '12 at 20:49
  • 4
    @Sean: IMHO, the PHP documentation is really quite bad at describing the nuances of core language features. But that is, perhaps, because so many ad-hoc special cases are baked into the language... – Oliver Charlesworth Feb 28 '13 at 8:43

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