CODEWITHSUNDEEP
haskell
Maqruis1
Maqruis1

Reputation: 167

Haskell: Couldn't match expected type ‘(a, b0)’ with actual type ‘(a, b, h)’

My code: 

g :: (a, b, h) -> (c, d, i) -> ((a, c))
g x y = ((fst x, fst y))

Why does it print the following errors?: 

ax.hs:11:15: error:
    • Couldn't match expected type ‘(a, b0)’
                  with actual type ‘(a, b, h)’
    • In the first argument of ‘fst’, namely ‘x’
      In the expression: fst x
      In the expression: ((fst x, fst y))
    • Relevant bindings include
        x :: (a, b, h) (bound at ax.hs:11:3)
        g :: (a, b, h) -> (c, d, i) -> (a, c) (bound at ax.hs:11:1)
   |
11 | g x y = ((fst x, fst y))
   |               ^

ax.hs:11:22: error:
    • Couldn't match expected type ‘(c, b1)’
                  with actual type ‘(c, d, i)’
    • In the first argument of ‘fst’, namely ‘y’
      In the expression: fst y
      In the expression: ((fst x, fst y))
    • Relevant bindings include
        y :: (c, d, i) (bound at ax.hs:11:5)
        g :: (a, b, h) -> (c, d, i) -> (a, c) (bound at ax.hs:11:1)
   |
11 | g x y = ((fst x, fst y))

Please help. I'm not sure why this is happening. How do I fix it? Thank you

Upvotes: 1

Views: 478

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477608

fst :: (a, b) -> a only works for 2-tuples, not tuples with a different arity than 2.

You can use pattern matching to obtain the first item of the 3-tuple, and personally I would use pattern matching for the second tuple as well, since this is - in my opinion - a more transparent syntax:

g :: (a, b, h) -> (c, d, i) -> (a, c)
g (x, _, _) (y, _, _) = (x, y)

Note that ((a, c)) is the same as (a, c), so we can remove the extra parenthesis.

Upvotes: 7

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