我有这两种表达方式：
foldr (-) 0 . map (uncurry (*)) $ coords 5 7
foldr (-) 0 . map (uncurry (*)) (coords 5 7)
(1) 工作打印出结果，但 (2) 有错误说：
<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)
谁能告诉我这两者有什么区别？谢谢。
有一个更简单的例子：
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的绑定比(.)。固定水平为($)解决这个问题：
id . id $ "Example" = (id . id) $ "Example"
= (id . id) "Example"
然而，随着(...)，函数应用程序获胜，您最终使用(.)以非函数作为第二个参数：
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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