isset() check flagged as undefined variable

[paulschreiber]

I would not expect and error from this code:

21 if ( ! isset( $content_width ) ) {
22 	$content_width = 944;
23 }

FILE: functions.php
-------------------------------------------------------------------------------------------------------------------------
FOUND 0 ERRORS AND 1 WARNING AFFECTING 1 LINE
-------------------------------------------------------------------------------------------------------------------------
 21 | WARNING | Variable $content_width is undefined.
    |         | (VariableAnalysis.CodeAnalysis.VariableAnalysis.UndefinedVariable)
-------------------------------------------------------------------------------------------------------------------------

[sirbrillig]

Ah, good point! isset and empty should probably both be allowed to specify a variable that doesn't exist.

[sirbrillig]

A tangent to this is that reading an undefined variable after a successful isset() or ! empty() check should not be reported as undefined. That is,

if (isset($undefinedVar)) { // no warning
    doSomething($undefinedVar); // also no warning
}

This is tricky and I'm not sure if I can do it though because while it's easy to see that in the above example $undefinedVar must be defined, what about this?

if ($override || isset($undefinedVar)) { // no warning
    doSomething($undefinedVar); // probably should still be a warning
}

It could get quite complex to tell if the variable is defined or not after the isset. And the logic necessary to do so is getting pretty close to static analysis territory, which is better handled by something like psalm or phpstan. So unfortunately I don't think we'll be able to add that here, but I thought it was worth mentioning (maybe you or someone else can come up with a more foolproof way to figure it out).

The example you gave is fine, though, since line 22 is a write rather than a read, and those are quite safe to use with an undefined variable.