[英]Haskell's function application operator ($) usage
I'm reading a piece by Bartosz Milewski wherein he defines the following function: 我正在读Bartosz Milewski撰写的一篇文章,其中他定义了以下功能:
instance Applicative Chan where
pure x = Chan (repeat x)
(Chan fs) <*> (Chan xs) = Chan (zipWith ($) fs xs)
Why is the function application operator in parenthesis? 为什么函数应用程序运算符在括号中? I understand this is normally done in order to use an infix function in prefix notation form, but I don't understand why, in this case, the function couldn't couldn't simply be expressed as Chan (zipWith $ fs xs)
, and wonder what the difference between the two is. 我理解这通常是为了在前缀表示形式中使用中缀函数,但我不明白为什么,在这种情况下,函数不能简单地表达为Chan (zipWith $ fs xs)
,并想知道两者之间的区别是什么。
(if you still need context, refer to the article ) (如果您仍需要上下文,请参阅文章 )
In this case, $
is being passed in to zipWith
. 在这种情况下, $
正被传递给zipWith
。 It's the same as writing 这和写作一样
zipWith (\ f x -> f x) fs xs
Without parentheses, it would have been equivalent to 没有括号,它本来就相当于
zipWith (fs xs)
which is not going to typecheck. 哪个不会出现问题。
An operator in parentheses behaves exactly like a normal identifier. 括号中的运算符与普通标识符完全相同。 With the following definition: 具有以下定义:
apply = ($)
the code could have looked like 代码可能看起来像
zipWith apply fs xs
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