cast.void

Code example

<?php declare(strict_types = 1);

$a = (void)(1 + 1);

Why is it reported?

The (void) cast (PHP 8.5+) can only be used as a standalone statement to explicitly discard a return value. It cannot be used within an expression because (void) does not produce a value -- assigning it, passing it as an argument, or nesting it inside another expression is not valid.

How to fix it

Use the (void) cast as a standalone statement instead of within an expression:

 <?php declare(strict_types = 1);

-$a = (void)(1 + 1);
+(void)(1 + 1);