为什么用这个表达式中的圆括号代替美元符号（$）会导致错误？ [英] Why does replacing the dollar sign ($) with parentheses in this expression leads to an error?

问题描述

我有这两个表达式：

I have these two expressions:

1. foldr（ - ）0。 map（uncurry（*））$ coords 5 7

foldr（ - ）0。地图（uncurry（*））（coords 5 7）

工作打印结果，但（2）有错误说：

The (1) works print out the result, but (2) have error says:

<interactive>:50:15:
    Couldn't match expected type ‘a -> t0 c’
                with actual type ‘[Integer]’
    Relevant bindings include
      it :: a -> c (bound at <interactive>:50:1)
    Possible cause: ‘map’ is applied to too many arguments
    In the second argument of ‘(.)’, namely
      ‘map (uncurry (*)) (coords 5 7)’
    In the expression: foldr (-) 0 . map (uncurry (*)) (coords 5 7)

有没有人能告诉我这两个？感谢。

Can any one tell me what's the difference between these two? Thanks.

推荐答案

有一个更简单的例子：

There's an easier example:

Prelude> id . id $ "Example"
"Example"
Prelude> id . id ("Example")
<interactive>:2:10:
    Couldn't match expected type ‘a -> c’ with actual type ‘[Char]’
    Relevant bindings include it :: a -> c (bound at <interactive>:2:1)
    In the first argument of ‘id’, namely ‘("Example")’
    In the second argument of ‘(.)’, namely ‘id ("Example")’
    In the expression: id . id ("Example")

问题是函数应用程序绑定比 （。）。 （$）的固定级别解决了这个问题：

The problem is that function application binds stronger than (.). The fixity level of ($) fixes this:

id . id $ "Example" = (id . id) $ "Example"
                    = (id . id) "Example"

例子=（id。id）$例子
=（id。id）

但是，使用（...）时，函数应用程序会赢，最终使用（。）作为第二个参数：

However, with (...), the function application wins and you end up using (.) with a non-function as second argument:

id . id ("Example") = id . id "Example"
                    = id . (id "Example") -- apply id
                    = id . ("Example")
                    = type error, since "Example" isn't a function

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