Question

假设我有以下列表

[((Int, Int), [[Char]], [[Char]], Bool, Bool, Bool)]
[(Int, Int, [Char], [Char], [Char], [Char], [Char], [Char], [Char], [Char])]

我知道为了能够打印它们，我需要将它们定义为新的数据类型，然后创建一个Show的实例，但是有什么方法可以避免它吗？

提前致谢。

Answer 1

让[Char]替换所有String以摆脱一些括号重载：

[((Int, Int), [String], [String], Bool, Bool, Bool)]

[(Int, Int, String, String, String, String, String, String, String, String)]

这些已经可以打印出来了：

Prelude> let x = replicate 2 ((1,2), ["a", "b"], ["c", "d"], True, False, True) :: [((Int,Int), [String], [String], Bool, Bool, Bool)]

Prelude> :type x   -- displays the type of x
x :: [((Int, Int), [String], [String], Bool, Bool, Bool)]

Prelude> print x
[((1,2),["a","b"],["c","d"],True,False,True),((1,2),["a","b"],["c","d"],True,False,True)]

和

Prelude> let x = replicate 2 (1,2,"a","b","c","d","e","f","g","h") :: [(Int,Int,String,String,String,String,String,String,String,String)]

Prelude> :type x
x :: [(Int,
       Int,
       String,
       String,
       String,
       String,
       String,
       String,
       String,
       String)]

Prelude> print x
[(1,2,"a","b","c","d","e","f","g","h"),(1,2,"a","b","c","d","e","f","g","h")]

Haskell：Deriving Show显示大元组列表

1 个答案: