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Home/ Questions/Q 637025
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T20:34:45+00:00 2026-05-13T20:34:45+00:00

If this works: Prelude Data.Char> map toUpper (sdfsd ++ dfgfdg) SDFSDDFGFDG Then why this

  • 0

If this works:

Prelude Data.Char> map toUpper ("sdfsd" ++ "dfgfdg")
"SDFSDDFGFDG"

Then why this doesn’t?

Prelude Data.Char> map toUpper . (++) "sdfsd" "dfgfdg"

<interactive>:1:14:
    Couldn't match expected type `a -> [Char]'
           against inferred type `[Char]'
    In the second argument of `(.)', namely `(++) "sdfsd" "dfgfdg"'
    In the expression: map toUpper . (++) "sdfsd" "dfgfdg"
    In the definition of `it': it = map toUpper . (++) "sdfsd" "dfgfdg"
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  1. Editorial Team
    Editorial Team
    2026-05-13T20:34:45+00:00Added an answer on May 13, 2026 at 8:34 pm
    map toUpper . (++) "sdfsd" "dfgfdg"
    

    is parsed as:

    (map toUpper) . ((++) "sdfsd" "dfgfdg")
    

    So basically you’re doing

    (map toUpper) . "sdfsddfgfdg"
    

    This does not work because the second argument to . needs to be a function, not a string.

    I assume you were trying to do something more like (map toUpper . (++)) "sdfsd" "dfgfdg". This also does not work because the return type of ++ is [a] -> [a] while the argument type of map toUpper is [a].

    The thing here is that while one might think of ++ as a function that takes two lists and returns a list, it really is a function that takes one list and then returns a function which takes another list and returns a list. To get what you want, you’d need to ++ into a function that takes a tuple of two lists and returns a list. That’s called uncurrying. The following works:

    map toUpper . (uncurry (++)) $ ("sdfsd", "dfgfdg")
    
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