Add is overload for free function predicate
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JohelEGP
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Sep 9, 2023
filipsajdak
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Current implementation of `is` does not handle generic functions as predicates.
The following case works:
```cpp
less_than_10_a: (i : int) -> bool = i < 10;
main: () = {
12 is (less_than_10_a);
}
```
It work also when function has generic return type
```cpp
less_than_10_b: (i : int) -> _ = i < 10;
main: () = {
12 is (less_than_10_b);
}
```
It does not work when function has generic argument.
The following case does not work:
```cpp
less_than_10_c: (i) -> _ = i < 10;
main: () = {
12 is (less_than_10_c); // error
}
```
This change introduce additional overload for `is` making free function
predicates acceptable by `is` (the function needs to return
`bool` or generic type). That makes the following cases to work:
```cpp
less_than_10_c: (i) -> _ = i < 10;
less_than_10_d: (i) -> bool = i < 10;
main: () = {
5 is (less_than_10_c); // works
2 is (less_than_10_d); // works
}
```
Functions that return other type then `bool` or `auto` do not work.
filipsajdak
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|
I am closing this one as it is covered by #701 |
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The current implementation of
isdoes not handle generic functions as predicates.The following case works:
It works also when the function has a generic return type:
But, it does not work when the function has a generic argument. The following case does not work:
This change introduces additional overload for
is, making the free function predicates acceptable byis(the function needs to returnboolor generic type). That makes the following cases work:Functions that return other types than
boolorautodo not work.All regression tests pass. Add a test for this case and update the regression test results.