Haskell - 如何写(。)f f =(\ x - > f(f x)) [英] Haskell - How to write (.) f f = (\x -> f (f x))
问题描述
我需要在GHCi上运行的模块上编写一个功能组合函数。这(经典的 fog(x)= f(g(x))
)运行:
<.c $ c>(。)fg =(\ x - > f(gx))。
当我尝试像这样写这个问题时出现问题
(。)ff =(\ x - > f(fx))。 (fof(x)= f(f(x)))
GHCi说:
f'的冲突定义
绑定在:Lab1.hs:27:9
Lab1.hs:27 :12
第27行:第9行出现,第27行:第27行再次出现f 。
为什么Haskell不能理解(。)ff =(\ x - > f(fx))$ c $在Haskell中,函数的参数必须具有唯一的名称。对另一个参数使用相同的名称是不允许的。这是因为
foo xy = ... === foo =(\x->(\y- > ...))
如果 y
替换为 x
,第二个 x
只会影响中的第一个。 ..
body:没有办法从那里引用第一个 x
。
您可以定义两次fx = f(fx)
:
lockquote
前奏曲>:t两次
两次::(t - > t) - > t - > t
前奏>两次(+1)4
6
另外, f(fx)=(。)ffx = join(。)fx
:
Prelude Control.Monad>:t join(。)
加入(。)::(b - > b) - > b - > b
join
在 Control.Monad
中定义。对于函数,它认为加入g x = g x x
。它也被称为 W combinator 。
例如 print $ join(。)(+1)4
prints 6 a>。
I need to write on a module to be run on GHCi, with a function composition to the same function. This (The classic fog(x) = f(g(x))
) runs:
(.) f g = (\x -> f (g x)).
The problem appears when I try to write it like this
(.) f f = (\x -> f (f x)). (fof(x) = f(f(x)))
GHCi says:
"Conflicting definitions for `f'
Bound at: Lab1.hs:27:9
Lab1.hs:27:12"
Line 27:9 appear on the first time f and line 27:12 appear f again.
Why doesn't Haskell understand (.) f f = (\x -> f (f x))
?
In Haskell, arguments to a function must have unique names. Using the same name for another argument is not allowed. This is because
foo x y = ... === foo = (\x-> (\y-> ...))
and if y
where replaced with x
, the second x
would just shadow the first inside the ...
body: there would be no way to reference the first x
from there.
You can just define twice f x = f (f x)
:
Prelude> :t twice
twice :: (t -> t) -> t -> t
Prelude> twice (+1) 4
6
Alternatively, f (f x) = (.) f f x = join (.) f x
:
Prelude Control.Monad> :t join (.)
join (.) :: (b -> b) -> b -> b
join
is defined in Control.Monad
. For functions, it holds that join g x = g x x
. It is also known as W combinator.
E.g. print $ join (.) (+1) 4
prints 6.
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