От: | achmed | https://www.linkedin.com/in/nail-achmedzhanov-9907188/ | |
Дата: | 15.10.09 06:16 | ||
Оценка: |
# let k (x:'a -> 'b) = x x;;
Characters 23-24:
let k (x:'a -> 'b) = x x;;
^
This expression has type 'a -> 'b but is here used with type 'a