A side question for a variable reference in the "PHP" Foreach Loop

$a = Array (' A ', ' B ', ' C '), foreach ($a as & $v) {}foreach ($a as $v) {}var_dump ($a);

Right now. Don't open your browser and guess. What is the result of the output?

Children's shoes, which are more familiar with references, may already be seen.  The correct answer is: Array (3) {[0]=> string (1) "A" [1]=> string (1) "B" [2]=> &string (1) "B"} is a,b,b. If you're guessing that's a,b,c. So for reference, you need to look at the relevant information: http://www.php.net/manual/zh/language.references.php

So why is it a,b,b? Let's take a step-by-step look at:

We know that when a foreach loop is executed on an array, it is implemented by moving the internal pointers to the arrays (for more details, you can read the PHP source code). So for the example in this article: when the Foreach loop ends, because $v is

Reference variables, so $v and $a [2] point to the same address space (shared variable values), and any subsequent modifications to $v will be reflected directly in the array $ A. We can add debug code to the example, we will be clear, for example, we in the second loop inside, plus var_dump ($a), testing each cycle when the value of a change:

$a = Array (' A ', ' B ', ' C '), foreach ($a as & $v) {}foreach ($a as $v) {var_dump ($a); echo "<br/>";} Var_dump ($a);

Run the code. The result is:

Draw a picture: you can see it more clearly: (the "$v pointing to $a[2]" is not accurate. It should be: $v and $a[2] point to the same place)

A few simple explanations for references:

1. A reference is similar to a pointer, but differs from a pointer.

For example, for references:

$a = "str"; $b = & $a;//<var class= "varname" ><var class= "varname" > $a </var></var> and <var class= "varname" ><var class= "varname" > $b </var></var> points to the same place

A simple one is as follows:

Then change the value of any element in $ A and $b at this time. The other value is changed as follows:

$a = "str"; $b = & $a; $b = "sssss"; echo $a;

2.unset will only delete the variable. Does not empty the memory space corresponding to the variable value: (This is different from the pointer)

$a = "str"; $b = & $a; unset ($b); Echo $a;

3. When a reference is passed as a function parameter, it can be changed internally by the function:

function change (& $a) {if (Is_array ($a)) {$a = array ();}} $test = Range (1,10), change ($test);p rint_r ($test);

Based on the above points, you should use the reference carefully during the encoding process. prevented from falling into inexplicable embarrassment.

PS: Do you understand? Try this question:

$a = range (1,3), foreach ($a as & $b) {    $b *= $b;} foreach ($a as $b) {    echo  $b;}   

Guess what the output is?

A side question for a variable reference in the "PHP" Foreach Loop