简体   繁体   English

无法使新数据类型Hashable

[英]Can't make a new data type Hashable

Attempt to compile the following with ghc 7.8.2 on OS X 10.9.3 尝试在OS X 10.9.3上使用ghc 7.8.2编译以下内容

import Data.Hashable (Hashable, hash)

data Edge v = Edge v v deriving (Show)

instance (Eq v) => Eq (Edge v) where
  Edge x1 x2 == Edge y1 y2 =
    x1 == y1 && x2 == y2 || x1 == y2 && x2 == y1

instance (Hashable v) => Hashable (Edge v) where
  hash (Edge x1 x2) = (hash x1) + (hash x2)

fails with 失败了

Could not deduce (hashable-1.2.1.0:Data.Hashable.Class.GHashable
                    (GHC.Generics.Rep (Edge v)))
  arising from a use of ‘hashable-1.2.1.0:Data.Hashable.Class.$gdmhashWithSalt’
from the context (Hashable v)
  bound by the instance declaration at src/MinCut.hs:12:10-42
In the expression:
  hashable-1.2.1.0:Data.Hashable.Class.$gdmhashWithSalt
In an equation for ‘hashWithSalt’:
    hashWithSalt
      = hashable-1.2.1.0:Data.Hashable.Class.$gdmhashWithSalt
In the instance declaration for ‘Hashable (Edge v)’

What is wrong? 怎么了?

The hackage docs for Data.Hashable states that the minimal implementation of Hashable is the function hashWithSalt — check out the documentation under the typeclass declaration ( class Hashable a where ). Data.Hashable的hackage文档声明Hashable的最小实现是函数hashWithSalt - 查看类型类声明下的文档( class Hashable a where )。

So if you change your function to hashWithSalt , everything should work: 因此,如果您将函数更改为hashWithSalt ,一切都应该工作:

instance (Hashable v) => Hashable (Edge v) where
  hashWithSalt s (Edge x1 x2) = s + (hash x1) + (hash x2)

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM