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

问题描述

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

输入的格式为 [Apple",15.00",5"]，我使用 Haskell 的 words 函数将其分解.

The input will be in the format ["Apple","15.00","5"] which I broke up using Haskell's words function.

我收到以下错误，我认为这是因为 makeGroceryItem 接受 Float 和 Int.

I get the following error which I think is because makeGroceryItem accepts a Float and an Int.

*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*

但是我如何分别制作 Float 和 Int 类型的 b 和 c ?

But how do I make b and c of type Float and Int, respectively?

推荐答案

read 可以将字符串解析为 float 和 int:

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

但问题 (1) 出在您的模式中:

But the problem (1) is in your pattern:

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

这里的 : 是一个(右结合)二元运算符，它将元素添加到列表中.元素的 RHS 必须是一个列表.因此，给定表达式 a:b:c，Haskell 将推断出以下类型:

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]

即c 将被认为是一个字符串列表.显然它不能被 read 或传递给任何需要 String 的函数.

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.

相反，你应该使用

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

如果列表必须正好有 3 个项目，或者

if the list must have exactly 3 items, or

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

如果 ≥3 个项目是可以接受的.

if ≥3 items is acceptable.

还有(2)，表达式

makeGroceryItem a read b read c

将被解释为带有 5 个参数的 makeGroceryItem，其中 2 个是 read 函数.您需要使用括号:

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)

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