If you're overloading ArrayObject, it's worth noting that while this method (when implemented by the parent) will return a reference, so code like $fakeArray['foobar']['hello'] = 1; will work like you expect.
However, when you overload the offsetGet method, you CANNOT define it as &offsetGet, so the above code falls out (because it returns the 'foobar' variable before you actually work with it).
This is something that the developers broke between 5.0 and 5.1, and was closed as bogus (http://bugs.php.net/bug.php?id=34783). So this is not a big, or question, or request, but just something worth noting.
ArrayObject::offsetGet
(PHP 5, PHP 7, PHP 8)
ArrayObject::offsetGet — Returns the value at the specified index
参数
-
key -
The index with the value.
返回值
The value at the specified index or null.
错误／异常
Produces an E_NOTICE error message when the specified
index does not exist.
示例
示例 #1 ArrayObject::offsetGet() example
<?php
$arrayobj = new ArrayObject(array('zero', 7, 'example'=>'e.g.'));
var_dump($arrayobj->offsetGet(1));
var_dump($arrayobj->offsetGet('example'));
var_dump($arrayobj->offsetExists('notfound'));
?>以上示例会输出：
int(7) string(4) "e.g." bool(false)
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User Contributed Notes 3 notes
Sam ¶
16 years ago
Jason ¶
8 years ago
It's worth noting that the PHP devs *did* fix the issue brought up in Sam's comment in 2013 (per the comment thread in his link here: https://bugs.php.net/bug.php?id=34783#1366088374), since at least PHP 5.3.8.
So, you *can* (and in most cases should) overload offsetGet with reference syntax to get expected functionality:
<?php
class myArrayType extends ArrayAccess {
public function &offsetGet($index) {
// ...
}
}
?>
Alex Andrienko ¶
15 years ago
Speaking of offsetGet() method overloading, be advised, that if you're iterating through Object via foreach, this method wouldn't be called. Iterator's current() method will be called instead.
