3 ms·
Haskell does it just fine with the HasField typeclass, but like everything Haskell, it's not always easy to find the information. type HasFooBar s = (HasFi
by pyrale 2y ago
Haskell does it just fine with the HasField typeclass, but like everything Haskell, it's not always easy to find the information.
type HasFooBar s = (HasField "_foo" s String, HasField "_bar" s String) => s