将字符串转换为整数/浮点在Haskell？ [英] Convert String to Integer/Float in Haskell?

问题描述

 数据GroceryItem = CartItem ItemName价格数量| StockItem项目名称价格数量
 
 makeGroceryItem :: String  - >浮动 - > Int  - > GroceryItem 
 makeGroceryItem name price quantity = CartItem name price quantity 
  

我想创建一个GroceryItem使用字符串或[String]

  createGroceryItem :: [String]  - > GroceryItem 
 createGroceryItem（a：b：c）= makeGroceryItem abc 
  

我用haskell中的单词函数分解了格式[Apple，15.00，5]。我得到这个错误，我认为是因为makeGroceryItem接受一个浮点数和一个整数。但是，我如何分别做b和c Float和Int？

  *在应用程序中输入错误
表达式：makeGroceryItem a读取b读取c 
 ** *术语：makeGroceryItem 
 ***类型：字符串 - >浮动 - > Int  - > GroceryItem 
 ***不符：a  - > b  - > c  - > d  - > e  - > f * 
  

解决方案

code>可以将字符串解析为float和int：

  Prelude> ：set + t 
 Prelude>阅读123.456:: Float 
 123.456 
 it :: Float 
 Prelude>阅读123456:: Int 
 123456 
 it :: Int 
  

但是问题（1）是在你的模式：

  createGroceryItem（a：b：c）= ... 
  

这里：是一个（右联合）将一个元素预先添加到列表中。元素的RHS必须是一个列表。因此，给定表达式 a：b：c ，Haskell会推断出以下类型：

  a :: String 
b :: String 
c :: [String] 
  

。即， c 将被认为是一个字符串列表。很明显，它不能是 read ，或者传递给任何需要字符串的函数。

p>

  createGroceryItem [a，b，c] = ... 
  

如果列表必须包含3个项目，或者

  createGroceryItem（a ：b：c：xs）= ... 
  

if≥3items are acceptable。

另外（2），表达式

  makeGroceryItem a read b read c 
  

会被解释为 makeGroceryItem 其中2个是 read 函数。你需要使用括号：

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ >

data GroceryItem = CartItem ItemName Price Quantity | StockItem ItemName Price Quantity

makeGroceryItem :: String -> Float -> Int -> GroceryItem
makeGroceryItem name price quantity = CartItem name price quantity  

I want to create a GroceryItem when using a String or [String]

createGroceryItem :: [String] -> GroceryItem
createGroceryItem (a:b:c) = makeGroceryItem a b c

The input will be in the format ["Apple","15.00","5"] which i broke up using words function in haskell. I get this error which i think is because the makeGroceryItem accepts a Float and an Int. But how do i make b and c Float and Int respectively?

*Type error in application
*** Expression     : makeGroceryItem a read b read c
*** Term           : makeGroceryItem
*** Type           : String -> Float -> Int -> GroceryItem
*** Does not match : a -> b -> c -> d -> e -> f*

解决方案

read can parse a string into float and int:

Prelude> :set +t
Prelude> read "123.456" :: Float
123.456
it :: Float
Prelude> read "123456" :: Int
123456
it :: Int

But the problem (1) is in your pattern:

createGroceryItem (a:b:c) = ...

Here : is a (right-associative) binary operator which prepends an element to a list. The RHS of an element must be a list. Therefore, given the expression a:b:c, Haskell will infer the following types:

a :: String
b :: String
c :: [String]

i.e. c will be thought as a list of strings. Obviously it can't be read or passed into any functions expecting a String.

Instead you should use

createGroceryItem [a, b, c] = ...

if the list must have exactly 3 items, or

createGroceryItem (a:b:c:xs) = ...

if ≥3 items is acceptable.

Also (2), the expression

makeGroceryItem a read b read c

will be interpreted as makeGroceryItem taking 5 arguments, 2 of which are the read function. You need to use parenthesis:

makeGroceryItem a (read b) (read c)

这篇关于将字符串转换为整数/浮点在Haskell？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！