PHP reference passing by value study notes_PHP tutorial

To use reference assignment in PHP, you only need to add an & before the original object. $a = &$b; In fact, a reference in PHP is two variables with different names pointing to the same value.

What is a quote

Quoting in PHP means accessing the same variable content with different names. This is not like a C pointer; instead, the reference is a symbol table alias. Note that in PHP, variable names and variable contents are different, so the same content can have different names. The closest analogy is Unix's filenames and the files themselves - the variable names are the directory entries, and the variable contents are the files themselves. References can be thought of as hardlinks in Unix file systems.

1: Variable reference

代码如下	复制代码
$a =100; $b = &$a; echo $b; //这里输出100 echo $a; //这里输出100 ，说明$a,和$b的值都是一百。 $b= 200; echo $a; //这里输出200 echo $b; //这里输出200,这就可以看出他们用的是同一个地址。改变一个，另一个也会跟着改变。 ?>

Two: Pass by value in the function.

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代码如下	复制代码
function main($a,$b){ $b= $a+100; return $b; } main(55,&$b); //这里的$b其实就是把它的内存地址传递给函数main中的$b参数，通过参数$b的改变而改变外面的$b的值。 echo $b; //这里会输出155， ?>

function main($a,$b){

$b= $a+100;

Return $b;

}

代码如下	复制代码
class club{ var $name="real madrid"; } $b=new club; $c=$b; echo $b->name;//这里输出real madrid echo $c->name;//这里输出real madrid $b->name="ronaldo"; echo $c->name;//这里输出ronaldo ?>

main(55,&$b); //The $b here actually passes its memory address to the $b parameter in the function main, and changes the value of the outer $b through the change of the parameter $b.

echo $b; //155 will be output here,
?>

代码如下	复制代码
$a = 'ronaldo' $b =&$a; unset ($a); ?>

Three: Passing objects by reference Object reference Unquote When you unset a reference, you just break the binding between the variable name and the variable's contents. This does not mean that the variable contents are destroyed. For example:

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$a = 'ronaldo'<🎜> $b =&$a;<🎜> unset ($a);<🎜> ?>

Won’t unset $b, just $a.

Example, pass by reference

test1.php

The code is as follows

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代码如下	复制代码
/** * 引用传递 以下内容可以通过引用传递： 变量，例如 foo($a) New 语句，例如 foo(new foobar()) 从函数中返回的引用，例如： */ function foo(&$var) { $var++; } $a=5; //合法 foo($a); foo(new stdClass()); //非法使用 function bar() // Note the missing & { $a = 5; return $a; } foo(bar()); // 自 PHP 5.0.5 起导致致命错误 foo($a = 5) // 表达式，不是变量 foo(5) // 导致致命错误 ?>

/**

* Pass by reference

代码如下

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function test(&$a) { $a=$a+100; } $b=1; echo $b;//输出１ test($b); //这里$b传递给函数的其实是$b的变量内容所处的内存地址，通过在函数里改变$a的值　就可以改变$b的值了 echo " ";   echo $b;//输出101   /***************************** * * 这里需要注意 call_user_func_array 后的参数是需要 & * * ****************************/       //上面的“ test($b); ” 中的$b前面不要加 & 符号，但是在函数“call_user_func_array”中，若要引用传参，就得需要 & 符号，如下代码所示：       function a(&$b){         $b++;     }     $c=0;     call_user_func_array('a',array(&$c));     echo $c; //输出 1 ?>

The following content can be passed by reference:

Variables, such as foo($a)

New statement, such as foo(new foobar())

The reference returned from the function, for example:

代码如下	复制代码
function &test() { static $b=0;//申明一个静态变量 $b=$b+1; echo $b; return $b; } $a=test();//这条语句会输出　$b的值　为１ $a=5; $a=test();//这条语句会输出　$b的值　为2 $a=&test();//这条语句会输出　$b的值　为3 这里将return $b中的　$b变量的内存地址与$a变量的内存地址　指向了同一个地方 $a=5; //已经改变了 return $b中的　$b变量的值 $a=test();//这条语句会输出　$b的值　为6 /**

*/<🎜> Function foo(&$var) <🎜> { <🎜>          $var++; <🎜> } <🎜> <🎜> $a=5; <🎜> //Legal <🎜> foo($a); <🎜> foo(new stdClass()); <🎜> //Illegal use <🎜> Function bar() // Note the missing & <🎜> { <🎜>         $a = 5; <🎜>         return $a; <🎜> } <🎜> foo(bar()); // Causes fatal error since PHP 5.0.5 <🎜> foo($a = 5) //Expression, not variable <🎜> foo(5) // Causes fatal error <🎜> <🎜> ?>

test2.php

The code is as follows

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function test(&$a) <🎜> { <🎜> $a=$a+100; <🎜> } <🎜> $b=1; <🎜> echo $b;//Output 1 <🎜> test($b); //What $b is passed to the function here is actually the memory address where the variable content of $b is located. By changing the value of $a in the function, the value of $b can be changed <🎜> <🎜> echo " "; echo $b;//Output 101 /***************************** * * It should be noted here that the parameters after call_user_func_array require & * * ****************************/ //Do not add the & symbol in front of $b in "test($b);" above, but in the function "call_user_func_array", if you want to pass parameters by reference, you need the & symbol, as shown in the following code: Function a(&$b){          $b++; } $c=0; ​ call_user_func_array('a',array(&$c)); echo $c; //Output 1 ?>

Return by reference Reference return is used when you want to use a function to find which variable the reference should be bound to. Don't use return references to increase performance, the engine is smart enough to optimize it itself. Only return references if there is a valid technical reason! To return a reference, use this syntax

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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. Here, the memory address of the $b variable in return $b and the memory address of the $a variable point to the same place <🎜> $a=5; //The value of the $b variable in return $b has been changed <🎜> <🎜> $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

Static variables are used here to let everyone understand the reference return of the function. In fact, the reference return of the function is mostly used in objects
*/
?>

To give an interesting example, I saw it on oschina:

( [0] => ABE

The code is as follows

代码如下	复制代码
$a = array('abe','ben','cam'); foreach ($a as $k=>&$n)     $n = strtoupper($n); foreach ($a as $k=>$n) // notice NO reference here!     echo "$nn"; print_r($a); ?>

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$a = array('abe','ben','cam');
foreach ($a as $k=>&$n)
$n = strtoupper($n);
foreach ($a as $k=>$n) // notice NO reference here!
echo "$nn";
print_r($a);
?>

will result in:

ABE
BEN
BEN
Array
(
[0] => ABE
[1] => BEN
[2] => BEN
)

Explanation: The loop in the second foreach is as follows:

Array
(
[0] => ABE
[1] => BEN

[2] => ABE

)

Array

[1] => BEN [2] => BEN

)

Array ( [0] => ABE [1] => BEN [2] => BEN ) Because there is no unset($n), it always points to the last element of the array. The first loop in the second foreach changes $n, that is, $a[2] to ABE, and the second loop I changed it to BEN, and it was BEN the third time.

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www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/632667.htmlTechArticleTo use reference assignment in php, you only need to add $b in front of the original object; in fact, the reference in php is two Variables with different names point to the same value. What is Quoting Quoting in PHP means using...