Haskell：列表理解：约束中的非类型变量参数：Num [t]

Question

When I am trying to run following function, I am getting error.

Prelude> let squareSum list = [result | (x, y, z) <- list, result <- x^2 + y^2 + z^2] 

<interactive>:4:5:
    Non type-variable argument in the constraint: Num [t]
    (Use FlexibleContexts to permit this)
    When checking that ‘squareSum’ has the inferred type
     squareSum :: forall t. Num [t] => [([t], [t], [t])] -> [t]

Can some one explain me, How to fix this? What is this error conveying exactly?

Answer 1

Original Question

You posted:

Prelude> let squareSum list = [result | (x, y, z) <- list, result <- x^2 + y^2 + z^2]

<interactive>:3:5:
    Non type-variable argument in the constraint: Num [t]
    (Use FlexibleContexts to permit this)
    When checking that ‘squareSum’ has the inferred type
      squareSum :: forall t. Num [t] => [([t], [t], [t])] -> [t]

This is coming from the inference that is the computation x^2 + y^2 + z^2 must be a list, due to using it as a source of values in a list comprehension ( result <- ... ). And if that is a list, then the mathematical operators are over list typed value, which means your initial variables, list must be a list of tuples of lists ( [([t],[t],[t])] ) and each list must in some way be valid numbers ( Num [t] ).

Comment Question

Prelude> let squareSum list = [ x^2 + y^2 + z^2 | (x, y, z) <- list]
Prelude> squareSum [1,2,3]

<interactive>:9:1:
    Non type-variable argument in the constraint: Num (t, t, t)
    (Use FlexibleContexts to permit this)
    When checking that ‘it’ has the inferred type
      it :: forall t. (Num t, Num (t, t, t)) => [t]

Now you say that the variable list contains tuples ( (x, y, z) <- list ) but then you define list as [1,2,3] . To satisfy both, the numeric literals of 1 , 2 and 3 must represent tuples, which is possible if you defined a class instance Num (t, t, t) .

What You Want

Carter already told you the solution, but you didn't apply it to a sensible list. How about we define a solution and give it an explicit type so there is less confusion!

Prelude> :{
Prelude| let squareSum :: [(Int,Int,Int)] -> [Int]
Prelude|     squareSum list = [ x^2 + y^2 + z^2 | (x, y, z) <- list]
Prelude| :}
Prelude> squareSum [(1,2,3), (4,5,6)]
[14,77]

Success! We provided two tuples and get two Int results, yay!

Answer 2

Following snippet solved my problem.

Prelude> let processData fun list = [y | x <- list, let y = fun x]
Prelude> let sumSquare (x, y, z) = x^2 + y^2 + z^2
Prelude> 
Prelude> processData sumSquare [(1, 2, 3), (4, 5, 6)]
[14,77]