Error Identifier: paramOut.type
Every error reported by PHPStan has an error identifier. Here’s a list of all error identifiers. In PHPStan Pro you can see the error identifier next to each error and filter errors by their identifiers.
Code example #
<?php declare(strict_types = 1);
/**
* @param-out int $p
*/
function foo(mixed &$p): void
{
$p = 1;
$p = 'str'; // ERROR: Parameter &$p @param-out type of function foo() expects int, string given.
}
Why is it reported? #
A by-reference parameter has a @param-out PHPDoc tag declaring the type that the parameter should have when the function returns. The value assigned to the parameter inside the function body does not match the declared @param-out type.
The @param-out tag is a contract with callers: it guarantees that after the function call, the referenced variable will have the specified type. Assigning a value of an incompatible type breaks this contract.
How to fix it #
Ensure the assigned value matches the declared @param-out type:
<?php declare(strict_types = 1);
/**
* @param-out int $p
*/
function foo(mixed &$p): void
{
$p = 1;
- $p = 'str';
+ $p = 2;
}
Or update the @param-out type to reflect the actual values being assigned:
<?php declare(strict_types = 1);
/**
- * @param-out int $p
+ * @param-out int|string $p
*/
function foo(mixed &$p): void
{
$p = 1;
$p = 'str';
}
How to ignore this error #
You can use the identifier paramOut.type to ignore this error using a comment:
// @phpstan-ignore paramOut.type
codeThatProducesTheError();You can also use only the identifier key to ignore all errors of the same type in your configuration file in the ignoreErrors parameter:
parameters:
ignoreErrors:
-
identifier: paramOut.type