Haskell let-in Construct

Question

为什么这两个函数的结果不相等？

mnr = [0,1,2,3,4,5,6]  :: [Int]
name = "Max Mustermann" :: String

t1 = ("p1",(length.take 2)mnr, (take 2.(let no name = name;in no))"No");
{- ("p1",2,"No") -}

t1' = ("p1",(length.take 2)mnr, (take 2.(let no n = name;in no))"No");
{- ("p1",2,"Ma") -}

这些函数的唯一区别是let中变量的名称.

最好的问候,Stefan

Answer 1

如果你打开-Wall,你会看到一个警告,t1暗示name现有的绑定name:

let no name = name
       ^^^^--- this one

阴影

name = "Max Mustermann" :: String

所以,name在函数内部的参数,使得功能一样id,而t2在name:

let no n = name

是在顶层定义的那个.