Bug #81330 [Nab]: Generator does not allow return null with implicit yield

Edit report at https://bugs.php.net/bug.php?id=81330&edit=1 ID: 81330 Updated by: cmb@php.net Reported by: greedy dot ivan at gmail dot com Summary: Generator does not allow return null with implicit yield Status: Not a bug Type: Bug Package: *General Issues Operating System: All PHP Version: 8.0.9 Assigned To: cmb Block user comment: N Private report: N New Comment: Actually, the point is that bar() is not a generator function at all, because there is no yield. You probably want something like <https://3v4l.org/hRSNS>. Previous Comments: ------------------------------------------------------------------------ [2021-08-04 12:25:36] greedy dot ivan at gmail dot com There are three options here. 1. There is an yield operator inside a function. Parser gets it and allow to return null as a valid end for generator. 2. There is no yield operator inside a function but there is a return instruction that returns Generator. This line works fine. 3. The same as 2, but we have a line, that returns null. It will throws Error, because Parser doesn't allow return an empty Generator such way. It doesn't get, that function return Generator at all, because there is no yield operation inside function itself. So, we must use some tricks to return empty Generator in this case, or use yield from func() instead for return func(). ------------------------------------------------------------------------ [2021-08-04 11:57:48] cmb@php.net That type check is done at runtime, not compile time. And there is no implicit yield. ------------------------------------------------------------------------ [2021-08-04 11:49:42] greedy dot ivan at gmail dot com It doesn't throw error on all other cases. It will throw Type error only in case, when yield is implicit and null is returned. ------------------------------------------------------------------------ [2021-08-04 11:39:32] cmb@php.net If the generator function does not necessarily return a Generator, you shouldn't declare it as such. ------------------------------------------------------------------------ [2021-08-04 11:29:25] greedy dot ivan at gmail dot com Description: ------------ If there is not explicit yield instruction in Generator, it will throw TypeError message when null is returned. https://3v4l.org/EpIjA Test script: --------------- <?php function foo($flag = false): \Generator { if ($flag) { return null; } yield from [1, 2]; } function bar($flag = false): \Generator { if ($flag) { return null; } return internal(); } function internal(): \Generator { yield from [1, 2]; } foreach (foo() as $v){} foreach (foo(true) as $v){} foreach (bar() as $v){} foreach (bar(true) as $v){} Expected result: ---------------- No error Actual result: -------------- Fatal error: Uncaught TypeError: bar(): Return value must be of type Generator, null returned in /in/EpIjA:15 ------------------------------------------------------------------------ -- Edit this bug report at https://bugs.php.net/bug.php?id=81330&edit=1