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
原文链接:https://www.f2er.com/php/136239.htmlhttp://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.