Question

我在Haskell中有这段代码拒绝编译：

data (Eq a, Num a, Show a) => Mat a = Mat {nexp :: Int, mat :: QT a}
 deriving (Eq, Show)

data (Eq a , Show a ) => QT a = C a | Q (QT a ) (QT a ) (QT a ) (QT a )
 deriving (Eq, Show)


cs:: (Num t) => Mat t -> [t]

cs(Mat nexp (Q a b c d)) =(css (nexp-1) a c)++(css (nexp-1) b d)
    where
        css 0 (C a) (C b) = (a-b):[]
        css nexp (Q a b c d) (Q e f g h) = (zipWith (+) (css (nexp-1) a c) (css (nexp-1) e g))++(zipWith (+)(css (nexp-1) b d) (css (nexp-1) f h))

我有这个错误：

Could not deduce (Show t) arising from a use of `Mat'
from the context (Num t)
bound by the type signature for cs:: Num t => Mat t -> [t]

我在网上搜索并发现了很多类似的问题，但似乎没有什么能接近我的问题。如何使这段代码有效？

Answer 1

数据类型的类约束don't mean anything.

请参阅this related question

删除数据类型的约束。

data Mat a = Mat {nexp :: Int, mat :: QT a}
    deriving (Eq, Show)

无法演绎（显示t）

1 个答案: