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От: |
nikov
|
http://www.linkedin.com/in/nikov |
Дата: | 01.06.09 12:15 | ||
Оценка: |
f, g :: a -> a
(f, g) = (id, id)
Couldn't match expected type `forall a. a -> a'
against inferred type `a -> a'
In the expression: id
In the expression: (id, id)
In a pattern binding: (f, g) = (id, id)