[BUG] Can't declare variable/parameter of type pointer to function #718

Activity

1. ntrel

   added

   bugSomething isn't workingSomething isn't working

   on Oct 2, 2023
2. JohelEGP commented on Oct 2, 2023

   

   JohelEGP

   on Oct 2, 2023

   Contributor

   More actions

   This is #711 and #343.
3. ntrel commented on Oct 2, 2023

   

   ntrel

   on Oct 2, 2023

   ContributorAuthor

   More actions

   Those use std::function, a template type, whereas this is about parsing a pointer to function type. Though it looks like #526 would fix this too.
4. JohelEGP commented on Oct 2, 2023

   

   JohelEGP

   on Oct 2, 2023

   Contributor

   More actions

   They're all about being able to name function types.
   To name a pointer to function type,
   you need to name a function type.
5. hsutter commented on Jul 27, 2024

   

   hsutter

   on Jul 27, 2024

   Owner

   More actions

   Thanks! Your code is now working with PR #1183:

   F: type == () -> int;
   f1: (a: *F)         = std::cout << a();
   f2: (a: *() -> int) = std::cout << a();
   
   g: () -> int = 42;
   
   main: () = {
       f1(g&);
       f2(g&);
   }
6. hsutter

   closed this as completedin #1183on Jul 27, 2024