I am trying to make a function that takes in any function as its first parameter and then returns that same function But I cannot figure out the type checking for it
I could do
function foo<A..., R...>(func: (A...) -> (R...)): (A...) -> (R...)
return func
end
But it does not support arguments and also it is too long and unreadable
Pls help! Any support would be appreciated
You could try letting the type solver infer it.
function foo<A..., R...>(func: (A...) -> (R...))
return func
end
Yeah but now it is not showing arguments
Does adding an “?” to the first R fix it? (in the previous code)
What do you mean?
I’m not sure if the type solver is able to do that.
Now it is just doing this
(I put an extra ? at the end of the Second R because it wouldn’t stop warning without it)
The only way to do it, is like this:
function a<Args..., Rets...>(f: (Args...) -> (Rets...)): (Args...) -> (Rets...)
return f
end
local b = a(function(hi: string): number
return 0
end)
b() -- (string): number
This method does support the types of the arguments.
You do lose all the argument names and documentation unfortunately.
That is intentional for now. Function types do not hold the information of the argument names, only the types. And that is enough in most cases.
That is just the same thing as this
Also I think I kind of found a solution
function foo<T> (func: T & (...any) -> (...any)): T
return func
end
But the only problem is that It autocompletes the members of the first parameter even if it is not a function
But At least it warns the user if the first parameter is not a function
I will not mark this as solution yet because there is Probably a better way to do this
