在使用模板 haskell 的 haskell 项目中，我尝试生成一个具有类型注释作为幻像类型的表达式。
一个简单的例子是这样的情况DataKinds and KindSignatures like:
{-# LANGUAGE DataKinds, KindSignatures #-}
data Foo = A | B
data GenMe (w :: Foo) = GenMe Int
[| $(generate some code) :: GenMe $(genType someCompileTimeData) |]
我怎样才能写一个函数，比如genType这样
genType :: Foo -> Q Type
提升只是提升保存编译时间的变量Foo价值？我不知道要使用哪个构造函数类型数据构造函数 https://hackage.haskell.org/package/template-haskell-2.12.0.0/docs/Language-Haskell-TH.html#t:Type制作数据类型。
有什么想法吗？谢谢！
解决这个问题的另一种方法是定义一个promote :: Exp -> Maybe Type函数，然后使用lift https://hackage.haskell.org/package/template-haskell-2.12.0.0/docs/Language-Haskell-TH-Syntax.html#v:lift on Foo.
-- | Takes the AST for an expression and tries to produce the corresponding
-- promoted type AST.
promote :: Exp -> Q Type
promote (VarE n) = fail ("Cannot promote variable " ++ show n)
promote (ConE n) = pure (PromotedT n)
promote (LitE l) = LitT <$> promoteLit l
promote (TupE es) = foldl AppT (PromotedTupleT (length es)) <$> (traverse promote es)
promote (ListE es) = foldr (\x xs -> AppT (AppT PromotedConsT x) xs) PromotedNilT <$> (traverse promote es)
promote (ParensE e) = ParensT <$> promote e
promote (AppE e1 e2) = AppT <$> promote e1 <*> promote e2
promote (InfixE (Just e1) e2 (Just e3)) = AppT <$> (AppT <$> promote e2 <*> promote e1) <*> promote e3
promote _ = fail "Either impossible to promote or unimplemented"
-- | Promote an expression literal to a type one
promoteLit :: Lit -> Q TyLit
promoteLit (StringL s) = pure (StrTyLit s)
promoteLit (IntegerL i) = pure (NumTyLit i)
promoteLit _ = fail "Expression literal cannot be promoted"
然后，我认为以下内容应该有效
ghci> :set -XDeriveLift -XDataKinds -XKindSignatures -XTemplateHaskell -XQuasiQuotes
ghci> data Foo = A | B deriving (Lift)
ghci> foo1 = A
ghci> data GenMe (w :: Foo) = GenMe Int
ghci> runQ [| GenMe 1 :: GenMe $(promote =<< lift foo1) |]
SigE (AppE (ConE Ghci5.GenMe) (LitE (IntegerL 1))) (AppT (ConT Ghci5.GenMe) (PromotedT Ghci3.A))
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