[ionfish]

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]

Yes, what does? Are you agreeing or disagreeing with my statement that currying is useless? I think you're agreeing.