PHP Reference Passing and Reference & Some Usage Introduction

Quotation in PHP is done using &. Now I will introduce to you some examples of usage and quotation issues and issues in PHP. Friends are welcome to enter for reference.

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.

What does a quote do
PHP’s references allow two variables to point to the same content.
When $a =& $b; $a and $b point to the same variable.
Tip: $a and $b are exactly the same here. It's not that $a points to $b or vice versa, but that $a and $b point to the same place.

You can pass a variable to a function by reference so that the function can modify the value of its argument. The syntax is as follows:

The code is as follows

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代码如下

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function foo(&$var)
{
$var++;
}

$a=5;
foo($a);
echo $a;

// 输出的是：6

function foo(&$var)
{

$var++;

}

代码如下

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$a = 100; //声明变量a
$b = &$a; //声明变量b,引用自变量a
echo "$a
";
echo "$b
";
$a++; //变量a自增1
echo "$a
";
echo "$b
";//查看变量b,也增加了1,说明使用的是同一存储单元
?>

程序运行结果：

100
100
101
101

$a=5;

foo($a);

echo $a;

//The output is: 6

代码如下	复制代码
$a = 20; $b = $a; $a = $a + 10; echo $a.' and '.$b; ?>

PHP quote character &

Regarding the role of PHP references (that is, adding the ampersand in front of variables, functions, objects, etc.), let’s first look at the following program.

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$a = 20; $b = &$a; $a = $a + 10; echo $a.' and '.$b; ?>

"; echo "$b
"; $a++; //Variable a increases by 1 echo "$a
"; echo "$b
";//Check variable b, it also increased by 1, indicating that the same storage unit is used ?> Program execution result: 100 100 101 101

Many people misunderstand that references in PHP are the same as pointers in C. In fact, they are not, and they are very different. Except for the pointers in C language that do not need to be explicitly declared during the array transfer process, other points need to be defined using *. However, the pointer to address (similar to a pointer) function in PHP is not implemented by the user himself, but is implemented by the Zend core. Yes, the reference in PHP adopts the principle of "copy-on-write", that is, unless a write operation occurs, variables or objects pointing to the same address will not be copied. php defaults to passing by value:

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$a = 20;<🎜> $b = $a;<🎜> $a = $a + 10; <🎜> echo $a.' and '.$b; <🎜> ?>

Program execution result: 30 and 20 If you want to pass it as an address, you need to add &, that is:

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$a = 20;<🎜> $b = &$a; <🎜> $a = $a + 10; <🎜> echo $a.' and '.$b; <🎜> ?>

Program execution result:

That is to say, & passes the address of $a to $b. In this case, the two variables now share a memory storage area, which means that their values are the same.

The same syntax can be used in functions, which return references, and in the new operator:

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代码如下

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view sourceprint?
1

2 $bar =& new fooclass();

3 $foo =& find_var($bar);

4 ?>

view sourceprint?

代码如下	复制代码
function foo(&$val1, $val2) { $val1 += 1; $val2 += 1; } $a=5; $b=10; foo($a,$b); echo $a; echo $b; ?>

2 $bar =& new fooclass();

3 $foo =& find_var($bar);

4 ?> The second thing a reference does is pass a variable by reference. This is accomplished by creating a local variable within the function that references the same content in the calling scope. To put it simply, the parameter of a function is a reference to a local variable. Here’s another example:

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function foo(&$val1, $val2) {

$val1 += 1;

代码如下

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$a = 1;
$b = &$a; //$a与$b指向同一内容
$b = 2;
echo $b; //2
echo $a; //2
传递引用
引用传递很简单，就是一个「&」符号，比如：

function foo(&$a) {
$a = 2;
}

$b = 1;
foo($b);
echo $b; //2

$val2 += 1;

}
$a=5;

$b=10;

代码如下

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class foo {
public $value = 42;

public function &getValue() { // 需要一个"&"
return $this->value;
}
}

$obj = new foo;
$myValue = &$obj->getValue(); // 还需要一个"&"，$myValue是对class foo中的$value的引用
$obj->value = 2; // 修改对象的$value属性
echo $myValue; // 输出2，$myValue与class foo中的$value值相同

foo($a,$b); echo $a; echo $b; ?> To run this code, you pass two parameters to the function, one is a reference to the content of $a, and the other is the value of $b. After executing this function, it is found that the content of $a has changed, while the content of $b is No changes. PHP references and misunderstandings References in PHP can be understood as aliases for variables. Since PHP variable names are stored in the symbol table (symbol table) and variable contents are stored in the heap, references use different symbol names in the symbol table to access the same storage content, which is the same as in the Unix file system. hardlink is the same concept, such as:

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$a = 1;<🎜> $b = &$a; //$a and $b point to the same content<🎜> $b = 2;<🎜> echo $b; //2<🎜> echo $a; //2<🎜> Pass reference <🎜> Passing by reference is very simple, it is just an "&" symbol, such as: <🎜> <🎜> function foo(&$a) {<🎜> $a = 2;<🎜> }<🎜> <🎜>$b = 1;<🎜> foo($b);<🎜> echo $b; //2<🎜>

<🎜>Return reference<🎜> In most cases, there is no need to return a reference to improve performance. The zend engine will optimize it by itself. However, if you must return a reference, you can return a reference in the following way: <🎜>

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<🎜> class foo {<🎜> Public $value = 42;<🎜> <🎜> public function &getValue() { // Requires a "&"<🎜> return $this->value; } } $obj = new foo; $myValue = &$obj->getValue(); // One more "&" is needed, $myValue is a reference to $value in class foo $obj->value = 2; // Modify the $value attribute of the object echo $myValue; // Output 2, $myValue is the same as $value in class foo

Differences from pointers
References are very similar to pointers, but they are not pointers. See the following code:

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代码如下	复制代码
$a = 0; $b = &a; echo $a; //0 unset($b); echo $a; //0

$a = 0;

$b = &a;

echo $a; //0

代码如下

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#include
int main(int argc, char const *argv[]) {
int a = 0;
int* b = &a;

    printf("%in", a); //0
    free(b);
    printf("%in", a); //*** error for object 0x7fff6350da08: pointer being freed was not allocated
}

unset($b); echo $a; //0 Since $b is just an alias of $a, even if $b is released, $a has no effect, but this is not the case for pointers. See the following code:

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#include int main(int argc, char const argv[]) { int a = 0; int b = &a; printf("%in", a); //0 free(b); Printf("%in", a); //*** error for object 0x7fff6350da08: pointer being freed was not allocated }

Since b is a pointer to a, after releasing the memory of b, an error will occur when accessing a again. This clearly illustrates the difference between PHP references and C pointers.

Objects and References
When using objects in PHP, everyone is always told that "objects are passed by reference". In fact, this is a misunderstanding. The object variable of PHP stores an identifier of the object. When passing the object, it is actually the identifier that is passed, not the reference. See the following code:

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代码如下

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$a = new A;
$b = $a;
$b->testA = 2;

/*
* 此时$a,$b的关系：
*        +-----------+      +-----------------+
* $a --> | object id | ---> | object(Class A) |
*        +-----------+      +-----------------+
*                               ^
*        +-----------+          |
* $b --> | object id | ---------+
*        +-----------+
*
*
*/

$c = new B;
$a = $c;
$a->testB = "Changed Class B";

/*
* 此时$a,$b,$c的关系：
*        +-----------+      +-----------------+
* $b --> | object id | ---> | object(Class A) |
*        +-----------+      +-----------------+
*
*        +------------+
* $a --> | object id2 | -------------+
*        +------------+              |
*                                    v
*        +------------+      +-----------------+
* $c --> | object id2 | ---> | object(Class B) |
*        +------------+      +-----------------+
*/

echo "object a: "; var_dump($a); //["testB"]=> string(15) "Changed Class B"
echo "object b: "; var_dump($b); //["testA"] => int(2)
echo "object c: "; var_dump($c); //["testB"]=> string(15) "Changed Class B"

$a = new A;<🎜> $b = $a; <🎜> $b->testA = 2; /* * The relationship between $a and $b at this time: * * $a --> | object id | ---> | object(Class A) | * * * * $b --> | object id | ---------+ * * * */ $c = new B; $a = $c; $a->testB = "Changed Class B"; /* * The relationship between $a, $b, $c at this time: * * $b --> | object id | ---> | object(Class A) | * * * * $a --> | object id2 | -------------+ * * * * $c --> | object id2 | ---> | object(Class B) | * */ echo "object a: "; var_dump($a); //["testB"]=> string(15) "Changed Class B" echo "object b: "; var_dump($b); //["testA"] => int(2) echo "object c: "; var_dump($c); //["testB"]=> string(15) "Changed Class B"

If the object is passed by reference, then the output content of $a, $b, $c should be the same, but in fact the result is not like this. Look at the following example of passing objects by reference:

$aa = new A;
$bb = &$aa; // Quote
$bb->testA = 2;

The code is as follows

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$aa = new A;
$bb = &$aa; // 引用
$bb->testA = 2;

/*
* 此时$aa, $bb的关系：
*
*         +-----------+      +-----------------+
* $bb --> | object id | ---> | object(Class A) |
*         +-----------+      +-----------------+
*              ^
*              |
* $aa ---------+
*
*
*/

$cc = new B;
$aa = $cc;
$aa->testB = "Changed Class B";

/*
* 此时$aa, $bb, $cc的关系：
*
*         +-----------+      +-----------------+
*         | object id | ---> | object(Class A) |
*         +-----------+      +-----------------+
*
* $bb ---->-----+
*               |
* $aa ---->-----+
*               |
*               v
*         +------------+
*         | object id2 | --------------+
*         +------------+               |
*                                      v
*         +------------+      +-----------------+
* $cc --> | object id2 | ---> | object(Class B) |
*         +------------+      +-----------------+
*/

echo "object aa: "; var_dump($aa); //["testB"]=>string(15) "Changed Class B"
echo "object bb: "; var_dump($bb); //["testB"]=>string(15) "Changed Class B"
echo "object cc: "; var_dump($cc); //["testB"]=>string(15) "Changed Class B"

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/*
* The relationship between $aa and $bb at this time:
*
*** * $bb --> | object id | ---> | object(Class A) |
* +-----------+ * ^ * * $aa ---------+
*
*
*/

$cc = new B;
$aa = $cc;
$aa->testB = "Changed Class B";

/*
* At this time, the relationship between $aa, $bb, $cc:
*
*                 +-----------+                                                                    * | object id | ---> | object(Class A) |
*                 +-----------+                                                                    *                                                                         * $bb ---->-----+
*                                                                     * $aa ---->-----+
*                                                                          *             v
* * | object id2 | ---------------+
*** *                                                                 v
* * $cc --> | object id2 | ---> | object(Class B) |
* */ echo "object aa: "; var_dump($aa); //["testB"]=>string(15) "Changed Class B"
echo "object bb: "; var_dump($bb); //["testB"]=>string(15) "Changed Class B"
echo "object cc: "; var_dump($cc); //["testB"]=>string(15) "Changed Class B" At this time, the contents of $aa, $bb, and $cc are exactly the same, so it can be seen that the object is not passed by reference. We need to get rid of this misunderstanding as soon as possible.

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PHP Reference Passing and Reference & Some Usage Introduction_PHP Tutorial