CODEWITHSUNDEEP
haskelldictionarypermutation
user1819047
user1819047

Reputation: 667

Haskell -- mapping function to permutation

I defined function listToNumber as below:

listToNumber = foldl1 (\acc xs -> acc*10 + xs)

It works fine when only supplied with one numeric list, for example:

listToNumber [1,2,3,4] = 1234
map listToNumber [[1,2,3,4], [5,4,3,2]] = [1234,5432]

However, the following returns error message:

map listToNumber permutations[1..3]

can someone explain please?

P.S. the error msg is as below:

Couldn't match expected type `[t1] -> t0' with actual type `[b0]'
The function `map' is applied to three arguments,
but its type `([b0] -> b0) -> [[b0]] -> [b0]' has only two
In the expression: map listToNumber permutations [1 .. 3]
In an equation for `it':
    it = map listToNumber permutations [1 .. 3]

Upvotes: 3

Views: 379

Answers (2)

Vladimir
Vladimir

Reputation: 7475

Another variant: map listToNumber $ permutations [1..3].

Upvotes: 0

jev
jev

Reputation: 2013

try map listToNumber (permutations [1 .. 3])

in ghci you can check the type of a function or an expression with :t

> :t map
> map :: (a -> b) -> [a] -> [b]

im means map requires a function and a list and returns a list, but in map listToNumber permutations [1 .. 3] you try to pass two functions and a list to map (since function application associates to the left).

Upvotes: 7

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