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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T21:51:19+00:00 2026-05-18T21:51:19+00:00

I was reading here , and I noticed that, for example, if I have

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I was reading here, and I noticed that, for example, if I have the following function definitions:

a :: Integer->Integer->Integer
b :: Integer->Bool

The following expression is invalid:

(b . a) 2 3

It’s quite strange that the functions of the composition must have only one parameter.

Is this restriction because some problem in defining the most generic one in Haskell or have some other reason?

I’m new to Haskell, so I’m asking maybe useless questions.

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  1. Editorial Team
    Editorial Team
    2026-05-18T21:51:20+00:00Added an answer on May 18, 2026 at 9:51 pm

    When you do a 2 3, you’re not applying a to 2 arguments. You’re actually applying a to it’s only argument, resulting in a function, then take that function and apply it to 3. So you actually do 2 applications. So in a sense, what you have is not equivalent to this:

    a :: (Integer, Integer) -> Integer
    b :: Integer -> Integer
    (b . a) (2, 3)
    

    You could’ve done this, btw

    (b . a 2) 3
    
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