哈斯克尔：GHC会对此进行优化吗？

Question

GHC能否简化id=（\（a，b）-

更复杂的情况呢：

id (Just x) = Just x
id Nothing = Nothing

map f (Just x) = Just (f x)
map _ Nothing  = Nothing

GHC将简化id。将映射到映射中？

我试图使用简单的beta缩减，但由于糟糕的模式匹配，这些术语看起来是不可缩减的。

因此，我很好奇GHC的优化技术如何处理这个问题。

Answer 1

您可以通过使用-ddump siml运行ghc来询问这些问题。这将导致ghc转储其编译程序的“核心”代码。Core是介于解释Haskell代码的编译器部分和将该代码转换为机器代码的编译器部分之间的中间语言。

当我用-O2-ddump siml编译下面的代码时，结果让我吃惊。

tupid1 :: (a, b) -> (a, b)
tupid1 = (\(a, b) -> (a, b))

tupid2 :: (a, b) -> (a, b)
tupid2 = (\(a, b) -> (a, b)) . (\(a, b) -> (a, b))

tupid1的结果核心生成了一个新的专用标识函数。

-- RHS size: {terms: 4, types: 7, coercions: 0}
tupid1 :: forall a_aqo b_aqp. (a_aqo, b_aqp) -> (a_aqo, b_aqp)
[GblId,
 Arity=1,
 Caf=NoCafRefs,
 Str=DmdType <S,1*U(U,U)>m,
 Unf=Unf{Src=InlineStable, TopLvl=True, Value=True, ConLike=True,
         WorkFree=True, Expandable=True,
         Guidance=ALWAYS_IF(arity=1,unsat_ok=True,boring_ok=True)
         Tmpl= \ (@ a_ayd)
                 (@ b_aye)
                 (ds_dIl [Occ=Once] :: (a_ayd, b_aye)) ->
                 ds_dIl}]
tupid1 = \ (@ a_ayd) (@ b_aye) (ds_dIl :: (a_ayd, b_aye)) -> ds_dIl

在内核中，函数的多态类型参数表示为显式参数tupid1为签名中的两个类型变量a和b获取其中两个类型参数，分别命名为a和b。它还接受一个术语ds_-dIl，该术语具有这两种类型的元组类型（ds_-dIl:：（a_-ayd，b_-aye）），并返回它而不进行修改。

令人惊讶的结果是tupid2。。。

-- RHS size: {terms: 1, types: 0, coercions: 0}
tupid2 :: forall a_aqm b_aqn. (a_aqm, b_aqn) -> (a_aqm, b_aqn)
[GblId,
 Arity=1,
 Caf=NoCafRefs,
 Str=DmdType <S,1*U(U,U)>m,
 Unf=Unf{Src=InlineStable, TopLvl=True, Value=True, ConLike=True,
         WorkFree=True, Expandable=True,
         Guidance=ALWAYS_IF(arity=1,unsat_ok=True,boring_ok=True)
         Tmpl= \ (@ a_axZ) (@ b_ay0) (x_aIw [Occ=Once] :: (a_axZ, b_ay0)) ->
                 x_aIw}]
tupid2 = tupid1

... 哪个ghc简化为tupid1！它是如何推断的超出了我的直接知识或发现能力。

的标识示例可能

maybeid :: Maybe a -> Maybe a
maybeid (Just x) = Just x
maybeid Nothing = Nothing

也简化为无模式匹配的标识函数

-- RHS size: {terms: 3, types: 4, coercions: 0}
maybeid :: forall a_aqn. Maybe a_aqn -> Maybe a_aqn
[GblId,
 Arity=1,
 Caf=NoCafRefs,
 Str=DmdType <S,1*U>,
 Unf=Unf{Src=InlineStable, TopLvl=True, Value=True, ConLike=True,
         WorkFree=True, Expandable=True,
         Guidance=ALWAYS_IF(arity=1,unsat_ok=True,boring_ok=True)
         Tmpl= \ (@ a_aqI) (ds_dIq [Occ=Once] :: Maybe a_aqI) -> ds_dIq}]
maybeid = \ (@ a_aqI) (ds_dIq :: Maybe a_aqI) -> ds_dIq

mapfor的核心可能对这个问题不感兴趣

maybemap :: (a -> b) -> Maybe a -> Maybe b
maybemap f (Just x) = Just (f x)
maybemap _ Nothing = Nothing

但是如果它是由maybeid

maybeidmap :: (a -> b) -> Maybe a -> Maybe b
maybeidmap f = maybeid . maybemap f

ghc将其简化为maybemap

-- RHS size: {terms: 1, types: 0, coercions: 0}
maybeidmap
  :: forall a_aqp b_aqq.
     (a_aqp -> b_aqq) -> Maybe a_aqp -> Maybe b_aqq
[GblId,
 Arity=2,
 Caf=NoCafRefs,
 Str=DmdType <L,1*C1(U)><S,1*U>,
 Unf=Unf{Src=InlineStable, TopLvl=True, Value=True, ConLike=True,
         WorkFree=True, Expandable=True,
         Guidance=ALWAYS_IF(arity=0,unsat_ok=True,boring_ok=True)
         Tmpl= maybemap}]
maybeidmap = maybemap

如果id由f组成，它也会做同样的事情。

maybemapid :: (a -> b) -> Maybe a -> Maybe b
maybemapid f = maybemap (id . f)

删除带有标识函数的合成，整个函数简化为maybemap

-- RHS size: {terms: 1, types: 0, coercions: 0}
maybemapid
  :: forall a_aqq b_aqr.
     (a_aqq -> b_aqr) -> Maybe a_aqq -> Maybe b_aqr
[GblId,
 Arity=2,
 Caf=NoCafRefs,
 Str=DmdType <L,1*C1(U)><S,1*U>,
 Unf=Unf{Src=InlineStable, TopLvl=True, Value=True, ConLike=True,
         WorkFree=True, Expandable=True,
         Guidance=ALWAYS_IF(arity=2,unsat_ok=True,boring_ok=False)
         Tmpl= \ (@ a_ar2)
                 (@ b_ar3)
                 (f_aqL [Occ=Once!] :: a_ar2 -> b_ar3)
                 (eta_B1 [Occ=Once!] :: Maybe a_ar2) ->
                 case eta_B1 of _ [Occ=Dead] {
                   Nothing -> GHC.Base.Nothing @ b_ar3;
                   Just x_aqJ [Occ=Once] -> GHC.Base.Just @ b_ar3 (f_aqL x_aqJ)
                 }}]
maybemapid = maybemap