4 ms·
Yes, it does. This is true in Haskell too, of course. subtract :: (Int, Int) -> Int subtract (n, m) = n - m fiveMinus = (curry subtract) 5
by ionfish 14y ago
Yes, it does. This is true in Haskell too, of course.
subtract :: (Int, Int) -> Int
subtract (n, m) = n - m
fiveMinus = (curry subtract) 5
minusFive = ???
subtract' = \n m -> n - m
fiveMinus = \m -> 5 - m
minusFive = \m -> m - 5
- tiglionabbit 14y agoYes, what does? Are you agreeing or disagreeing with my statement that currying is useless? I think you're agreeing.