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Error Identifier: generator.void

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);

/**
 * @return \Generator<int, int, void, void>
 */
function numbers(): \Generator
{
	yield 1;
	$value = yield 2;
}

Why is it reported? #

The generator declares void as its send type (the TSend template parameter of Generator), meaning that no value is sent back into the generator via Generator::send(). Using the result of a yield expression when the send type is void will always produce null, which indicates a logic error – the code appears to expect a value to be sent into the generator, but the type declaration says otherwise.

How to fix it #

If the generator needs to receive values via send(), declare the correct send type:

 /**
- * @return \Generator<int, int, void, void>
+ * @return \Generator<int, int, string, void>
  */
 function numbers(): \Generator
 {
 	yield 1;
 	$value = yield 2;
 }

If the generator does not need to receive values, remove the unused assignment:

 /**
  * @return \Generator<int, int, void, void>
  */
 function numbers(): \Generator
 {
 	yield 1;
-	$value = yield 2;
+	yield 2;
 }

How to ignore this error #

You can use the identifier generator.void to ignore this error using a comment:

// @phpstan-ignore generator.void
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: generator.void

Rules that report this error #

  • PHPStan\Rules\Generators\YieldFromTypeRule [1]
  • PHPStan\Rules\Generators\YieldTypeRule [1]

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