哈斯克爾：推導顯示大元組

的名單假設我有以下列出哈斯克爾：推導顯示大元組

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

我知道，爲了能夠打印他們，我需要將其定義爲一個新的數據類型，然後進行顯示的一個實例，但有什麼辦法可以避免它？

在此先感謝。

來源

2013-04-02 Kudayar Pirimbaev

元組你應該開始使用記錄，否則你會*忘記什麼是什麼。 –

是的，我知道，但由於它是一個任務，它有一些限制 –

讓我們全部更換[Char]通過String擺脫一些支架超載：

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

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

這些已經可以打印：

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")]

來源

2013-04-02 08:00:18

錯誤 - 表達式中的語法錯誤（意外輸入結束） –

@KudayarPirimbaev在哪裏？ –

let x = replicate ... –

哈斯克爾：推導顯示大元組

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