<?php
$a = array('a', 'b', 'c', 'd');
foreach ($a as &$v) { }
foreach ($a as $v) { }
print_r($a);
?>
I think it's a normal program but this is the output I am getting:
Array
(
[0] => a
[1] => b
[2] => c
[3] => c
)
Can some开发者_运维技巧one please explain this to me?
This is well-documented PHP behaviour See the warning on the foreach page of php.net
Warning
Reference of a $value and the last array element remain even after the foreach loop. It is recommended to destroy it by unset().
$a = array('a', 'b', 'c', 'd');
foreach ($a as &$v) { }
unset($v);
foreach ($a as $v) { }
print_r($a);
EDIT
Attempt at a step-by-step guide to what is actually happening here
$a = array('a', 'b', 'c', 'd');
foreach ($a as &$v) { } // 1st iteration $v is a reference to $a[0] ('a')
foreach ($a as &$v) { } // 2nd iteration $v is a reference to $a[1] ('b')
foreach ($a as &$v) { } // 3rd iteration $v is a reference to $a[2] ('c')
foreach ($a as &$v) { } // 4th iteration $v is a reference to $a[3] ('d')
// At the end of the foreach loop,
// $v is still a reference to $a[3] ('d')
foreach ($a as $v) { } // 1st iteration $v (still a reference to $a[3])
// is set to a value of $a[0] ('a').
// Because it is a reference to $a[3],
// it sets $a[3] to 'a'.
foreach ($a as $v) { } // 2nd iteration $v (still a reference to $a[3])
// is set to a value of $a[1] ('b').
// Because it is a reference to $a[3],
// it sets $a[3] to 'b'.
foreach ($a as $v) { } // 3rd iteration $v (still a reference to $a[3])
// is set to a value of $a[2] ('c').
// Because it is a reference to $a[3],
// it sets $a[3] to 'c'.
foreach ($a as $v) { } // 4th iteration $v (still a reference to $a[3])
// is set to a value of $a[3] ('c' since
// the last iteration).
// Because it is a reference to $a[3],
// it sets $a[3] to 'c'.
The first
foreach loop does not make any change to the array, just as we would expect.
However, it does cause $v
to be assigned a reference to each of $a
’s elements,
so that, by the time the first loop is over, $v
is, in fact, a reference to $a[2]
.
As soon as the second loop starts, $v
is now assigned the value of each
element. However, $v
is already a reference to $a[2];
therefore, any value
assigned to it will be copied automatically into the last element of the array!
Thus, during the first iteration, $a[2]
will become zero, then one, and then
one again, being effectively copied on to itself. To solve this problem, you
should always unset the variables you use in your by-reference foreach
loops—or, better yet, avoid using the former altogether.
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