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[英]Will GHC optimize a*a*a*a*a*a to (a*a*a)*(a*a*a)?
[英]Haskell : Will GHC optimize this?
GHC可以将id = (\\(a, b) -> (a, b)).(\\(a, b) -> (a, b))
简化为id = \\(a, b) -> (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会简化id . map
id . map
到map
?
我尝试使用普通的beta缩减,但看起来这些术语是不可简化的,因为讨厌的模式匹配。
因此,我很好奇GHC的优化技术是如何处理的。
您可以通过使用-ddump-simpl
运行它来询问ghc的这些问题。 这将导致ghc转储它编译程序的“核心”代码。 Core是编译器的一部分介于Haskell代码之间的中间语言,也是编译器将该代码转换为机器代码的部分。
当我使用-O2 -ddump-simpl
编译以下内容时,结果让我感到惊讶。
tupid1 :: (a, b) -> (a, b)
tupid1 = (\(a, b) -> (a, b))
tupid2 :: (a, b) -> (a, b)
tupid2 = (\(a, b) -> (a, b)) . (\(a, b) -> (a, b))
由此产生的tupid1
核心构成了一个新的专用身份函数。
-- 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
在核心中,函数的多态类型参数表示为显式参数。 tupid1
为其签名中的两个类型变量a
和b
采用其中两个类型参数,名为a_ayd
和b_aye
。 它还需要一个术语ds_dIl
,它具有这两种类型的元组类型( ds_dIl :: (a_ayd, b_aye)
)并且不加修改地返回它。
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
! 如何推断这是超出我的直接知识或发现的能力。
Maybe
的标识示例
maybeid :: Maybe a -> Maybe a
maybeid (Just x) = Just x
maybeid Nothing = Nothing
也简化为没有模式匹配的身份函数
-- 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
对于这个问题, Maybe
map
核心并不有意思
maybemap :: (a -> b) -> Maybe a -> Maybe b
maybemap f (Just x) = Just (f x)
maybemap _ Nothing = Nothing
但如果它是由maybeid
组成的
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)
删除具有身份功能的组合,整个功能简化为可能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
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