Subject: bug#59140: 29.0.50; iter-yield from lambda

From: Michael Heerdegen <[email-masked]>
Date: Sat, 16 Sep 2023 03:56:35 +0200

Max Brieiev <[email-masked]> writes:

> For example, this:
>
>     (iter-defun my-generator ()
>       (funcall (lambda () (iter-yield 5))))
>
> would be expanded by iter-defun macro into this:
>
>     (...
>       (let ((gen (iter-lambda () (iter-yield 5))))
>         (iter-next (funcall gen))))
>
> Does it make sense?

Does it?  Isn't the `let' expression equivalent to just `5'?  With other
words: you don't yield from an outside generator, as far as I
understand (or am I wrong? what's the content of your "..."?).

Michael.