Haskell - fast naming of projection functions

Question

Suppose I want to define a function f in terms of some other predefined function g as follows:

f :: Int -> Int -> Int
f 2 b = g b
f _ _ = 1

That is, I want to define the projection, f(2,_) : Int->Int to be the same as g(_) : Int->Int. Gladly, Haskell has first class functions, and so definitions like the following squarePlusOne are valid and standard:

plusOne :: Int -> Int
plusOne i = i+1
square :: Int -> Int
square i = i*i
squarePlusOne :: Int -> Int
squarePlusOne = plusOne . Square

With Haskell's currying (ie. f takes just one Int as input and returns an (Int->Int) typed function), I am surprised I cannot write

f 2 = g

Why not? Or is there some other syntax?

Oops! I've found that my question was totally founded by a different issue. — Owen, Apr 05 '19 at 06:56

score 4 · Answer 1 · answered Apr 05 '19 at 07:10

4

Indeed, writing f 2 = g is a valid way to define f. However, when defining functions in this way, remember you must define the whole function with the same pattern signature. That is, you may not exhaust your function by writing

f 2 = g
f i j = i+j

Instead, this may be achieved like so:

f 2 = g
f i = (\j-> i+j)

answered Apr 05 '19 at 07:10

Owen

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surely, to match the OP's original definition, the section line should be `f _ = \_ -> 1`? (Oh sorry, I didn't realise you *were* the OP until after I commented :) ) – Robin Zigmond Apr 05 '19 at 08:04
That's also just `f i = (i +)`. – chepner Apr 05 '19 at 14:54

chepner · Answer 2 · 2019-04-05T14:53:41.830

2

You can use the const function, which creates a function that ignores its argument to return a fixed value.

f :: Int -> Int -> Int
f 2 = g            -- f 2 x = g x
f _ = const 1      -- f _ x = const 1 x == (\_ -> 1) x == 1

edited Apr 05 '19 at 14:53

answered Apr 05 '19 at 14:47

chepner

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Haskell - fast naming of projection functions

2 Answers2