我有以下代码段：
@H_403_1@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.
