Haskell:列表理解：约束中的非类型变量参数：Num[t]

当我试图运行下面的函数时，我得到了一个错误

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]

Prelude>let squareSum list=[result |（x，y，z）[t]

---

有人能给我解释一下，如何解决这个问题吗？这个错误到底是什么？

下面的片段解决了我的问题

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]

Prelude>let processData fun list=[y | x let sumSquare（x，y，z）=x^2+y^2+z^2
序曲>
前奏曲>过程数据sumSquare[（1,2,3）、（4,5,6）]
[14,77]

原始问题 您发布了：

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]

---

现在你说变量

list

包含元组（

（x，y，z）可能是因为你试图从一个和中提取
结果
——我想你想要
let squareSum list=[x^2+y^2+z^2 |（x，y，z）谢谢Carsten。我试过了，但仍然面临同样的问题。序曲>let squareSum list=[x^2+y^2+z^2 |（x，y，z）squareSum[1，2，3]：33:1：约束中的非类型变量参数：Num（t，t，t）（使用flexibleContext允许此操作）在检查“it”是否具有推断类型it:：forall t.（Num t，Num（t，t，t））=>[（1,2,3），（4,5,6）]
使用
让平方和xs=sum[x^2|x
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]

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]