[英]Confusing function application and function composition in Haskell
The operation 手术
(filter (`notElem` "'\"").[(1,'a','%',"yes")])
gives an error. 给出一个错误。 How can apply this filter on that list properly? 如何将此过滤器正确地应用于该列表?
You've got a couple of serious problems. 您遇到了几个严重的问题。 First, your syntax is wacky ( .
definitely shouldn't be there). 首先,您的语法很古怪( .
绝对不应该在那里)。 But the bigger problem is that what you're trying to filter is of the type [(Int,Char,Char,[Char])]
(that is, a list containing a 4-tuple). 但更大的问题是,您要过滤的内容是[(Int,Char,Char,[Char])]
(即包含4个元组的列表)。
And your list has only one element, which is (1,'a','%',"yes")
. 并且您的列表只有一个元素,即(1,'a','%',"yes")
。 So filtering that is useless anyway. 因此过滤仍然无济于事。 When function you provide for filtering must be of type a -> Boolean
, where a
is the type of all the elements of the list. 当您提供过滤的函数必须是a -> Boolean
类型时,其中a
是列表中所有元素的类型。
Seems like you wanted some sort of wonky heterogenous list or something. 似乎您想要某种类型的异类列表或其他内容。
The .
的.
operator in Haskell is function composition -- it composes two functions together. Haskell中的operator是函数组合-它将两个函数组合在一起。
So your code, 所以你的代码
(`notElem` "'\"") . [(1,'a','%',"yes")]
looks like the composition of the notElem
function and some list. 看起来像notElem
函数和一些列表的组成。 That's just wrong. 那是错误的。
Remove the .
删除.
, and make sure to show
the list first: ,并确保首先show
该列表:
> filter (`notElem` "'\"") (show [(1,'a','%',"yes")])
"[(1,a,%,yes)]"
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