Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Kind inference #1157

Open
dhil opened this issue Jul 22, 2022 · 0 comments
Open

Kind inference #1157

dhil opened this issue Jul 22, 2022 · 0 comments
Milestone

Comments

@dhil
Copy link
Member

dhil commented Jul 22, 2022

In the absence of a kind annotation Links assumes that a given type variable has kind Type. Consequently, one must explicitly declare the kinds of type variables with kinds other than Type in type aliases, and one must use the appropriate explicit syntax for these type variables in type applications, e.g.

typename Id(a, e::Eff) = (a) ~e~> a;

sig id : Id(a, { |e})
fun id(x) { x }

In this example, the necessary kind information is easily inferable. Therefore it would be convenient to be able to write the following instead

typename Id(a, e) = (a) ~e~> a;

sig id : Id(a, e)
fun id(x) { x }
@dhil dhil added this to the 0.9.8 milestone Jul 22, 2022
@dhil dhil modified the milestones: 0.9.8, 0.9.9 Nov 27, 2023
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Projects
None yet
Development

No branches or pull requests

1 participant