Questions about php references

//Return by reference

<code>function &testReturn(){  
    static $b = 1;  
    $b += 2;  
    return $b;  
}  
$a = &testReturn();  
$a = 8;  
$c = &testReturn();
$c = 12;
$d = testReturn();  
//echo $d;  

function &cuitReturn(){
    $a = 2;
    return $a;
}
$cr = &cuitReturn();
//echo $cr;
$cr = 4;
$cr1 = cuitReturn();
echo $cr1;
</code>

The second function changes the value of the assigned variable, $cr = 4; why does the return value of the function not change?

Reply content:

//Return by reference

<code>function &testReturn(){  
    static $b = 1;  
    $b += 2;  
    return $b;  
}  
$a = &testReturn();  
$a = 8;  
$c = &testReturn();
$c = 12;
$d = testReturn();  
//echo $d;  

function &cuitReturn(){
    $a = 2;
    return $a;
}
$cr = &cuitReturn();
//echo $cr;
$cr = 4;
$cr1 = cuitReturn();
echo $cr1;
</code>

The second function changes the value of the assigned variable, $cr = 4; why does the return value of the function not change?

The code of your first function is actually like this, because $b is a static variable, so it will not be released after the function is executed.

...    //省略代码
$c = &$b;
$c = 12;    //此处$b为12
$d = testReturn();    //$b+2
echo $d; //当然是14而不是7

But in the second function $a is a local variable. After the function is executed, the memory of this variable is released.

First of all, we need to make it clear whether the calling function returns a reference. The function name must be preceded by &, and the assignment statement must be preceded by &. So, in the title of the question, $cr1 = cuitReturn();is actually not a reference.

Back to what the questioner said, why the return value has not changed? It’s because the $a in the function cuitReturn is a local variable, and it is not static, so it is released after the function returns, $cr = &cuitReturn();It is equivalent to referencing a local variable, If this is placed in C++, it will cause big trouble...This means that the pointer points to unknown memory,But the PHP engine should handle it, so$rc The reference to $a is invalid