[英]Haskell: Point-free style
Why does the first one fails while the latter one succeeds in compilation? 为什么第一个失败而后一个成功编译?
I expect foo
and foo'
are equivalent, that is, foo'
is just a point-free function of foo
: 我希望foo
和foo'
是等价的,即, foo'
仅仅是一个免费的点对点功能foo
:
foo :: [a] -> [a] -> [(a,a)]
foo = map id . zip
foo' :: [a] -> [a] -> [(a,a)]
foo' a b = map id $ zip a b
But foo
fails with following error: 但是foo
失败,并显示以下错误:
Couldn't match type ‘[b0] -> [(a, b0)]’ with ‘[b]’
Expected type: [a] -> [b]
Actual type: [a] -> [b0] -> [(a, b0)]
Relevant bindings include
foo :: [a] -> [b] (bound at <interactive>:26:5)
Probable cause: ‘zip’ is applied to too few arguments
In the second argument of ‘(.)’, namely ‘zip’
In the expression: map id . zip
Any comments will be appreciated. 任何意见将不胜感激。
If you look at the definition/type signature of (.)
you see that its arguments are functions with a single parameter. 如果查看(.)
的定义/类型签名,则会看到其参数是具有单个参数的函数。 But zip
has two thus you have to supply at least one parameter to make it equivalent 但是zip
有两个,因此您必须提供至少一个参数以使其等效
foo a = map id . zip a
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