我怎样才能做到这一点
assert toList 1 2 3 == [1, 2, 3]
我在看
https://www.haskell.org/haskellwiki/Varargs
http://chris-taylor.github.io/blog/2013/03/01/how-haskell-printf-works/
http://paczesiowa.blogspot.de/2010/03/polyvariadic-primefib-problem.html
但是我还没有掌握它（对Haskell初级的学习还很多）
到目前为止我的尝试
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE UndecidableInstances #-}
{-# LANGUAGE OverlappingInstances #-}
class ToVector r where
process :: Int -> r
instance ToVector Int where
process = id
instance ToVector r => ToVector (Int -> r) where
process s = \v -> process v
对此表示欢迎
*Main> process 1 2 3
<interactive>:158:1:
Could not deduce (ToVector (a0 -> a1 -> t))
arising from the ambiguity check for ‘it’
from the context (ToVector (a -> a2 -> t), Num a2, Num a)
bound by the inferred type for ‘it’:
(ToVector (a -> a2 -> t), Num a2, Num a) => t
at <interactive>:158:1-13
The type variables ‘a0’, ‘a1’ are ambiguous
When checking that ‘it’
has the inferred type ‘forall a a1 t.
(ToVector (a -> a1 -> t), Num a1, Num a) =>
t’
Probable cause: the inferred type is ambiguous
*Main>
非常欢迎任何帮助和解释
这里的示例演示了如何使用您的代码。问题（由@Cubic正确识别）是多态性。特别是，您可以在示例代码中看到使用类型调用方法：
main :: IO ()
main = do printAll 5 "Mary" "had" "a" "little" "lamb" 4.2
printAll 4 3 5
您需要执行以下操作：
foo :: Int
foo = process (1 :: Int) (2 :: Int) (3 :: Int)
这指定的类型process，即to Int -> Int -> Int -> Int。您还需要指定数字的类型（与上面的示例不同），因为您仅为Ints创建了一个实例，而它1可以是任何Num类型。
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我来说两句