What does PHP custom function reference return? Detailed explanation of reference return instance

ReferenceReturn

This is what the manual says: Reference return is used when you want to use a function to find which the reference should be bound to. When variable is above. 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

When you want to bind the return reference of a function to a variable, PHP allows you to do this:

function &returns_reference()
{
  static $someref = 0;
  $someref++;
  return $someref;
}
 
$newref = &returns_reference();//引用返回,相当于 $newref = &$someref;
echo $newref; //1
//phpfensi.com
$notref = returns_reference(); //值传递的是副本
$newref = 100;
echo $notref; //2
 
$newref = 100;
echo returns_reference(); //101

It can be seen that if you want the function to return a reference, you must declare it in the function When assigning values, bring the & operator.

The same is true for methods in classes:

class foo {
  public $value = 0;
 
  public function &getValue() {
    return $this->value;
  }
}
 
$obj = new foo;
$myValue = &$obj->getValue(); // $myValue is a reference to $obj->value, which is 42.
$obj->value = 2;
echo $myValue;

Some simple Example

Look at the simple example below and try to understand the reference return.

<?php
function &test()
{
 // 声明一个静态变量
  static $b = 0;
  $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
$a = 5;
$a = test(); //这条语句会输出　$b的值　为6
?>
//程序运行结果：
1
2
3
6

Although the function declaration method isfunction &test() This way, But what we get from the function call in this way $a = test() is not actually a function reference return, which is no different from an ordinary function call. PHP stipulates that what is obtained through $a = &test() is the reference return of the function.

Using the above example to explain, $a = test() Calling a function in this way only assigns the value of the function to $a, and any changes made to $a will It will not affect $b in the function.

And when calling the function through $a = &test() , its function is to compare the memory address of the $b variable in return $b with the $a variable The memory address points to the same place. That is to say, the effect equivalent to this is produced ($a=&$b), so changing the value of $a also changes the value of $b.

So after executing

$a = &test();

$a = 5;

, the value of $b becomes 5.

Another program example to deepen understanding:

<?php
/*
** 值传递和引用传递，值传递传递的是值的一个复本，引用传递传递的是值指向的内存地址
*/
// 函数的引用，定义时也要加上 &
function &func($a,$b){ 
 // 这里为了更直观看到效果，定义一个静态变量
 static $result = 0;  
 $result+=$a+$b;
 echo $result.&#39;<br />&#39;;
 return $result;
}
 
$a = $b = 10;
// PHP里这样写函数的引用调用，和调用普通函数没有区别（只是将函数的返回值复制给$c这个变量，$c做任何改变不会影响上面函数中的$result）
// 要记住：PHP里的函数引用定义及调用都要在函数名前加上 &
$c = func($a,$b); 
// 第一次执行func()，其静态变量$result的值变为 20(10+10)
// 改变$c的值，不会对下面一行语句产生影响
$c = 666; 
// 第二次执行func()，其静态变量$result的值变为 40(20+10+10)
$c = func($a,$b);
echo &#39;<hr />&#39;;
// 这样才是PHP中引用函数的调用方式
$d = &func($a,$b); 
// 第三次执行func()，其静态变量$result的值变为 40(40+10+10)
$d = 888;
// 第四次执行func()，其静态变量$result的值变为 908(888+10+10)
$d = func($a,$b);
?>

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