php容易犯错的10个地方
原文地址：
http://www.toptal.com/php/10-most-common-mistakes-php-programmers-make
译文地址：http://codecloud.net/php-2056.html
Common Mistake #1: Leaving dangling array references after foreach loops
Not sure how to use foreach loops in PHP? Using references in foreach loops can be useful if you want to operate on each element in the array that you are iterating over. For example:
<code>$arr = array(1, 2, 3, 4);foreach ($arr as &$value) { $value = $value * 2;}// $arr is now array(2, 4, 6, 8)</code>
The problem is that, if you’re not careful, this can also have some undesirable side effects and consequences. Specifically, in the above example, after the code is executed, $value will remain in scope and will hold a reference to the last element in the array. Subsequent operations involving $value could therefore unintentionally end up modifying the last element in the array.
The main thing to remember is that foreach does not create a scope. Thus, $value in the above example is areference within the top scope of the script. On each iteration foreach sets the reference to point to the next element of $array. After the loop completes, therefore, $value still points to the last element of $array and remains in scope.
Here’s an example of the kind of evasive and confusing bugs that this can lead to:
<code>$array = [1, 2, 3];echo implode(',', $array), "\n";foreach ($array as &$value) {} // by referenceecho implode(',', $array), "\n";foreach ($array as $value) {} // by value (i.e., copy)echo implode(',', $array), "\n";</code>
The above code will output the following:
<code>1,2,31,2,31,2,2</code>
No, that’s not a typo. The last value on the last line is indeed a 2, not a 3.
Why?
After going through the first foreach loop, $array remains unchanged but, as explained above, $value is left as a dangling reference to the last element in $array (since that foreach loop accessed $value by reference).
As a result, when we go through the second foreach loop, “weird stuff” appears to happen. Specifically, since $value %本文@来源gao@!dai!ma.com搞$$代^@码!网搞代gaodaima码is now being accessed by value (i.e., by copy), foreach copies each sequential $array element into $value in each step of the loop. As a result, here’s what happens during each step of the second foreachloop:
- Pass 1: Copies
$array[0](i.e., “1”) into$value(which is a reference to$array[2]), so$array[2]now equals 1. So$arraynow contains [1, 2, 1]. - Pass 2: Copies
$array[1](i.e., “2”) into$value(which is a reference to$array[2]), so$array[2]now equals 2. So$arraynow contains [1, 2, 2]. - Pass 3: Copies
$array[2](which now equals “2”) into$value(which is a reference to$array[2]), so$array[2]still equals 2. So$arraynow contains [1, 2, 2].
To still get the benefit of using references in foreach loops without running the risk of these kinds of problems, call unset() on the variable, immediately after the foreach loop, to remove the reference; e.g.:
<code>$arr = array(1, 2, 3, 4);foreach ($arr as &$value) { $value = $value * 2;}unset($value); // $value no longer references $arr[3]</code>
Common Mistake #2: Misunderstanding isset() behavior
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