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What does it mean to add a number in front of a function when defining it? _PHP Tutorial
- WBOYOriginal
- 2016-07-13 10:51:041236browse
What does it mean to add an ampersand in front of a function when defining it?
If so, please give an example. Thank you
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[url=http://www.111cn.cn/bbs/space.php?uid=66514]Link tag[img]http://www.111cn.cn/server/avatar.php?uid=66514&size=small[ /img][/url] Let me reveal the correct answer below. This is what I saw in the post of E Snail Children's Shoes on the forum. It accurately explains the function of adding an & in front of the function and the specific effect.
Function reference returns
Look at the code first and copy the PHP content to the clipboard
PHP code:
function &test()
{
static $b=0;//Declare a static variable
$b=$b+1;
echo $b;
return $b;
}
$a=test();//This statement will output that the value of $b is 1
$a=5;
$a=test();//This statement will output that the value of $b is 2
$a=&test();//This statement will output that the value of $b is 3
$a=5;
$a=test();//This statement will output that the value of $b is 6
Explain below:
In this way, $a=test(); actually does not get a reference return from the function. It is no different from an ordinary function call. As for the reason: This is the regulation of PHP
PHP stipulates that what is obtained through $a=&test(); is the reference return of the function
As for what is a reference return (the PHP manual says: Reference return is used when you want to use a function to find which variable the reference should be bound to.) This nonsense made me unable to understand it for a long time
To explain using the above example,
Calling a function using $a=test() only assigns the value of the function to $a, and any changes to $a will not affect $b
in the function.
When calling a function through $a=&test(), its function is to point the memory address of the $b variable in return $b and the memory address of the $a variable to the same place
That is to say, the effect equivalent to this is produced ($a=&b;), so changing the value of $a also changes the value of $b, so after executing
$a=&test();
$a=5;
From now on, the value of $b becomes 5