php - 为什么第一次调用没有E_NOTICE错误？

Question

我有以下代码片段:

 error_reporting(E_ALL | E_STRICT);

    function &getVal() {
       $data = [];

       return $data['hey'];
       //return $whatever; 
    }

    function getVal2() {
       $data = [];

       return $data['hey'];
    }

    var_dump(getVal());  // No E_NOTICE error is issued - why?
    var_dump(getVal2()); // E_NOTICE error is issued.

问题是:为什么第一次调用没有E_NOTICE错误？解释很可能是创建变量 $data['hey'] 以返回引用。但是，当 $data['hey'](或 $whatever，...)时不发出 E_NOTICE 错误似乎仍然是错误的未定义。

Answer 1

这是预期的行为

http://www.php.net/manual/en/language.references.whatdo.php#language.references.whatdo.assign

If you assign, pass, or return an undefined variable by reference, it will get created.

和一些相关的“错误”:

https://bugs.php.net/bug.php?id=30350

Ok, it appears that the element is created because we are attempting to return a reference to something that does not exist.

https://bugs.php.net/bug.php?id=27627

When you try to access a non-existant array element you effectively create it, hence the NULL entries in the array.