約束返回類型的上下文

Question

這裏是我的嘗試至今：

module Main where 

data FooT = One | Two deriving (Show, Read) 
{- 
That is what I want 
foo :: (Show a, Read a) => a 
foo = One 
-} 

--class Footable (Show a, Read a) => a where 
class Footable a where 
    --foo2 :: (Show a, Read a) => a 
    foo2 :: a 

instance Footable FooT where 
    foo2 = One 

-- test = print foo2 

我想測試編譯。我不認爲這個問題圍繞着普遍的量化而展開。 GHC說，一個是一個「嚴格的類型變量」 編輯（剛性類型的變量），但我真的不理解這是什麼。這個問題似乎正如我在我的評論@寫sepp2k它可能對生存型，但我無意中發現了一個奇怪的行爲可能與this

編輯

：

這確實編譯：

{-# LANGUAGE OverlappingInstances, FlexibleInstances, OverlappingInstances, 
UndecidableInstances, MonomorphismRestriction, PolymorphicComponents #-} 
{-# OPTIONS_GHC -fno-monomorphism-restriction #-} 

module Main where 

    class (Num a) => Numable a where 
     foo2 :: a 

    instance (Num a) => Numable a where 
     foo2 = 1 

    instance Numable Int where 
     foo2 = 2 

    instance Numable Integer where 
     foo2 = 3 

    --test = foo2 + foo2 -- This does NOT compile (ambiguous a) 
    test = (foo2::Integer) + foo2 --this works 

但這並不（`a」是剛性類型的變量消息）

{-# LANGUAGE OverlappingInstances, FlexibleInstances, OverlappingInstances, 
UndecidableInstances, MonomorphismRestriction, PolymorphicComponents #-} 
{-# OPTIONS_GHC -fno-monomorphism-restriction #-} 

module Main where 

    data FooT = One | Two deriving (Show, Read) 
    data BarT = Ten deriving (Show, Read) 

    class (Show a, Read a) => Footable a where 
     foo2 :: a 

    instance (Show a, Read a) => Footable a where 
     foo2 = Ten 

    instance Footable FooT where 
     foo2 = One 

    main = print foo2 

這是因爲1 ::（Num t）=> t。我可以像這樣定義一些東西（typeconstructor，consts dunno）嗎？

Answer 1

當我取消的test的定義，並嘗試編譯代碼時，我得到「曖昧類型變量」。沒有什麼嚴格的。要理解爲什麼這是不明確的考慮：

module Main where 

data FooT = One | Two deriving (Show, Read) 
data BarT = Three | Four deriving Show 

class Footable a where 
    foo2 :: a 

instance Footable FooT where 
    foo2 = One 

instance Footable BarT where 
    foo2 = Three 

main = print foo2 -- Should this print One or Three? 

當然，在你的代碼中有Footable只有一個實例，所以哈斯克爾在理論上可以推斷出你想要使用的定義foo2爲FooT，因爲這是唯一的實例在適用範圍。但是如果這樣做，該代碼將盡快爲您導入的情況來定義Footable的另一個實例模塊摔壞，所以Haskell沒有做到這一點。

要解決你的問題，你需要用它的類型來註釋foo2的，像這樣：

module Main where 

data FooT = One | Two deriving (Show, Read) 

class Footable a where 
    foo2 :: a 

instance Footable FooT where 
    foo2 = One 

main = print (foo2 :: FooT) 

要求所有Footables是顯示和閱讀的情況下，簡單地做：

class (Show a, Read a) => Footable a where 
    foo2 :: a 

像你這樣在你的評論中，但沒有再次在foo2的簽名中指定約束。

Answer 2

正如sepp2k說，GHC無法猜測foo2的返回類型。做它的約束（這是你的問題的標題）添加一個內聯類型的簽名。

測試=打印（foo2的::英尺）