Questions about PHP references

Reference returns

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;

The second function, changing the value of the assignment variable, $CR = 4; Why did the return value of the function not change?

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Reference returns

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;

The second function, changing the value of the assignment variable, $CR = 4; Why did the return value of the function not change?

Your first function, the code is actually like this, because $b it is a static variable, so the function will not be released after execution.

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

But the second function $a is a local variable, the function is finished, the memory of this variable is freed.

The first thing to be clear is whether the calling function returns a reference function name preceded by a &, and the assignment statement is preceded by A & so the title $cr1 = cuitReturn(); is actually not a reference to this point.

Back to the main said, why the return value is not changed, because cuitReturn the function $a is a local variable, and is not static, so the function is returned after the release, equivalent to a $cr = &cuitReturn(); reference to a local variable, which if placed in C + + will be a big event ... This means that the pointer is pointing to an unknown memory, but the PHP engine should be handled, so $rc the reference to it $a is invalid

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