php – 为什么第一次调用中没有E_NOTICE错误？

发布时间：2020-12-13 13:05:02 所属栏目：PHP教程来源：网络整理

导读：我有以下代码段： 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

我有以下代码段：

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：为什么第一次调用中没有E_NOTICE错误？解释很可能是创建变量$data [‘hey’]来返回引用.但是,当$data [‘hey’](或$whatever,…)未定义时,发出E_NOTICE错误仍然是错误的.

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.

还有一些相关的“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.

（编辑：李大同）

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