如何寫地圖/ fmap模擬（a-> b） - > F m a - > F m b

Question

好日/夜大家！ 我有型：

data FixItem m a = KeepItem|SkipItem|FixItem (m (a -> a)) 
fixItem f = FixItem $ pure f 

，我想寫功能mapFix :: (a -> b) -> FixItem m a -> FixItem m b。當我嘗試：

mapFix f SkipItem = SkipItem -- good 
mapFix f KeepItem = fixItem f -- error "rigid type"!!! 
mapFix f (FixItem mf) = FixItem $ pure (.) <*> (pure f) <*> mf -- too! 

所以，我得到錯誤：

• Couldn't match type ‘b’ with ‘a’ 
     ‘b’ is a rigid type variable bound by 
     the type signature for: 
      mapFix :: forall (m :: * -> *) a b. 
        Applicative m => 
        (a -> b) -> FixItem m a -> FixItem m b 
     at src/test.hs:235:11 
     ‘a’ is a rigid type variable bound by 
     the type signature for: 
      mapFix :: forall (m :: * -> *) a b. 
        Applicative m => 
        (a -> b) -> FixItem m a -> FixItem m b 
     at src/test.hs:235:11 
     Expected type: b -> b 
     Actual type: a -> b 
    • In the first argument of ‘fixItem’, namely ‘f’ 
     In the expression: fixItem f 
     In an equation for ‘mapFix’: mapFix f KeepItem = fixItem f 
    • Relevant bindings include 
     f :: a -> b (bound at src/test.hs:236:8) 
     mapFix :: (a -> b) -> FixItem m a -> FixItem m b 
      (bound at src/test.hs:236:1) 

如何寫mapFix或實現函子實例這種類型（FixItem修復a到a，不b，IE修復是a -> a ，而不是a -> b）？

Answer 1

您不能爲您的數據類型實施Functor類型類。這是因爲你的一個構造函數中有 a -> a。當你有功能時，你應該更加小心。但簡而言之，您將逆變類型變量a鍵入，因此您無法在此類型變量上實現Functor。

儘管您可以爲您的數據類型實施Invariant。因爲a在協變和逆變職位，您的數據類型是不變的函子。

可以幫助您：

Example of Invariant Functor?

What is a contravariant functor?

Some blog post