显示哈斯克尔的结果

发布于 2025-02-06 15:59:41 字数 183 浏览 0 评论 0原文

我是Haskell的新手。

如何返回（x1，x2）并从我的代码中打印出来？

qqq x
   | x < 0  x1 = mod (-x) 10 
   | 1 < x && x < 99 x1 = mod x 10
   | x2 = mod x 10

原文

I'm very new to haskell.

How can I return (x1,x2) and print it out from my code?

qqq x
   | x < 0  x1 = mod (-x) 10 
   | 1 < x && x < 99 x1 = mod x 10
   | x2 = mod x 10

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煞人兵器 2025-02-13 15:59:41

您以错误的方式使用守卫。您似乎将它们视为如果语句，则可以将其用于 atsignement 。在Haskell中，您不将分配值为变量，您 neclare 这些。您可以使用：

qqq :: Integral a => a -> (a, a)
qqq x
   | x < 0 = (mod (-x) 10, x2)
   | 1 < x && x < 99 = (mod x 10, x2)
  where x2 = mod x 10

因此，每个后卫在方程符号之前都有条件（=），在右侧，返回了2件托盘，将第一个项目返回x1 ，作为第二项x2。

您还应该实现x == 1和x＆gt; = 99的额外案例。

You are using guards the wrong way. You seem to see these as if statements, that you then can use for assignements. In Haskell, you do not assign values to a variable, you declare these. You can work with:

qqq :: Integral a => a -> (a, a)
qqq x
   | x < 0 = (mod (-x) 10, x2)
   | 1 < x && x < 99 = (mod x 10, x2)
  where x2 = mod x 10

Here each guard thus has a condition before the equation sign (=), and at the right side returns a 2-tuple with as first item an expression for x1, and as second item x2.

You should also implement extra case(s) for x == 1 and x >= 99, these are not covered by the two guards.

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