题
instance Monad (Either a) where
return = Left
fail = Right
Left x >>= f = f x
Right x >>= _ = Right x
此代码在“ baby.hs”中引起了可怕的汇编错误:
Prelude> :l baby
[1 of 1] Compiling Main ( baby.hs, interpreted )
baby.hs:2:18:
Couldn't match expected type `a1' against inferred type `a'
`a1' is a rigid type variable bound by
the type signature for `return' at <no location info>
`a' is a rigid type variable bound by
the instance declaration at baby.hs:1:23
In the expression: Left
In the definition of `return': return = Left
In the instance declaration for `Monad (Either a)'
baby.hs:3:16:
Couldn't match expected type `[Char]' against inferred type `a1'
`a1' is a rigid type variable bound by
the type signature for `fail' at <no location info>
Expected type: String
Inferred type: a1
In the expression: Right
In the definition of `fail': fail = Right
baby.hs:4:26:
Couldn't match expected type `a1' against inferred type `a'
`a1' is a rigid type variable bound by
the type signature for `>>=' at <no location info>
`a' is a rigid type variable bound by
the instance declaration at baby.hs:1:23
In the first argument of `f', namely `x'
In the expression: f x
In the definition of `>>=': Left x >>= f = f x
baby.hs:5:31:
Couldn't match expected type `b' against inferred type `a'
`b' is a rigid type variable bound by
the type signature for `>>=' at <no location info>
`a' is a rigid type variable bound by
the instance declaration at baby.hs:1:23
In the first argument of `Right', namely `x'
In the expression: Right x
In the definition of `>>=': Right x >>= _ = Right x
Failed, modules loaded: none.
为什么会发生这种情况?我该如何编译该代码?感谢您的帮助〜
我知道了。我调整了代码以查看它的编译:
instance Monad (Either a) where
return = Right
Left a >>= f = Left a
Right x >>= f = f x
它成功编译!但是...还有更多问题:
instance Monad (Either a)
使“一个'a monad和我得到'return =正确'...我怎么能得到'return = weled'?我已经尝试过,但失败了:
instance Monad (`Either` a) where
return = Left
Right a >>= f = Right a
Left x >>= f = f x
或:实例monad( x-> XA)
根本不编译!
解决方案
- 返回应该有类型
forall b. b -> Either a b
, 但是,左有类型forall c. a -> Either a c
. 。您可能想在这里。 fail
应该有类型forall b. String -> Either a b
, ,但是正确的类型forall b. b -> Either a b
, , 因此,如果b=String
这使得String -> Either a String
这不合适。>>=
应该有类型Either a b -> (b -> Either a c) -> Either a c
然而Right x >>= _ = Right x
总是返回类型的值Either a b
, , 不是Either a c
.Left x >>= f = f x
因为x有类型,所以不起作用a
, , 但f
有类型b -> c
.
其他提示
大多数混乱源于左右事实。仅考虑返回的类型,其单本类型的类型如下:
return :: (Monad m) => b -> m b
您正在尝试定义一个实例 m
= Either a
, ,因此返回应该具有类型:
return :: b -> Either a b
您将其定义为左侧,其中具有类型:
Left :: a -> Either a b
注意如何 ->
有所不同。
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