[
Т.е.
_::Int -> Int
Это одна дырка, а
sqrt:: Num a => a => a
sqrt = _a
Другая? Или я путаю?
Size: a a a
[
MK
Maybe _[
AA
AA
MK
MK
MK
[
DG
:t (.) (.) :: (a1 -> b -> c) -> a1 -> (a2 -> b) -> a2 -> _Выдаёт:
• Found type wildcard ‘_’ standing for ‘c1’Можно ли сделать так, чтобы <expression> стало равно "(a1 -> b -> c) -> a1 -> (a2 -> b) -> a2 -> с"
Where: ‘c1’ is a rigid type variable bound by
the inferred type of
<expression> :: (a4 -> b1 -> c1) -> a4 -> (a5 -> b1) -> a5 -> c1
...
AA
AA
AA
MK
AA
AA
MK
AA
[
