`isset` (and `??` operator) could should a notice if the inner expression is non-array

[mvorisek]

Description

The following code:

$oneDimensionalArray = [];
var_dump($oneDimensionalArray[0][1]);
var_dump($oneDimensionalArray[0][1] ?? '');

https://3v4l.org/1LMRe

Resulted in this output:

Warning: Undefined array key 0 in /in/1LMRe on line 4

Warning: Trying to access array offset on value of type null in /in/1LMRe on line 4
NULL
string(0) ""

But I expected this output instead:

Warning: Undefined array key 0 in /in/1LMRe on line 4

Warning: Trying to access array offset on value of type null in /in/1LMRe on line 4
NULL
Warning: Undefined array key 0 in /in/1LMRe on line 4

string(0) ""

PHP Version

8 - master

Operating System

any

[mvorisek]

isset should also not suppress such notice: https://3v4l.org/Nn6Hg

[bwoebi]

This is not a bug. It's the default behaviour of isset() to do a safe, warning-free access check (and ?? shares isset() semantics).

I.e. isset() does not only check the existence of a key or property, but also whether it can be dereferenced at all via the specific operator.

[bwoebi]

@iluuu1994 Oh, you retagged that just now. Well, feel free to reopen that as a feature if you consider it worthwhile pursuing. I think the current semantics make perfect sense.

Anyway, in any case, a change here would require a RFC.

[mvorisek]

Yes, please reopen, retitled and desc updated

[bwoebi]

@mvorisek your last case is explicitly allowed by isset docs:

 If multiple parameters are supplied then isset() will return true only if all of the parameters are considered set. Evaluation goes from left to right and stops as soon as an unset variable is encountered.