parameter.void

Code example

<?php declare(strict_types = 1);

function doFoo(void $param): void // ERROR: Parameter $param has typehint with void.
{
}

Why is it reported?

The void type cannot be used as a parameter type in PHP. The void type is only valid as a return type, where it indicates that the function does not return a value. Using void as a parameter type is a language-level error that will cause a fatal error at runtime.

How to fix it

Use the appropriate type for the parameter. If the parameter is intended to accept no meaningful value, consider whether the parameter is needed at all:

 <?php declare(strict_types = 1);

-function doFoo(void $param): void
+function doFoo(): void
 {
 }

If the parameter should accept any type including null, use mixed:

 <?php declare(strict_types = 1);

-function doFoo(void $param): void
+function doFoo(mixed $param): void
 {
 }