如何在Haskell中返回积分? [英] How to return an Integral in Haskell?

问题描述

我试图弄清楚Haskell，但是我对'Integral'有点迷恋.据我了解，Int和Integer都是整数.但是，如果我尝试编译这样的函数:

I'm trying to figure out Haskell, but I'm a bit stuck with 'Integral'. From what I gather, Int and Integer are both Integral. However if I try to compile a function like this:

lastNums :: Integral a => a -> a
lastNums a = read ( tail ( show a ) ) :: Integer

我知道

Could not deduce (a ~ Integer)
from the context (Integral a)

如何返回积分?

也可以说我必须坚持使用该功能签名.

Also lets say I have to stick to that function signature.

推荐答案

让我们用英语阅读此函数类型签名.

Let's read this function type signature in English.

lastNums :: Integral a => a -> a

这意味着让调用者选择任何整数类型.lastNums函数可以采用该类型的值并产生相同类型的另一个值."

This means that "Let the caller choose any integral type. The lastNums function can take a value of that type and produce another value of the same type."

但是，您的定义始终返回 Integer .根据类型签名，应该将决定留给调用者.

However, your definition always returns Integer. According to the type signature, it's supposed to leave that decision up to the caller.

最简单的解决方法:

lastNums :: Integer -> Integer
lastNums = read . tail . show

定义单态函数不会感到羞耻.不要仅仅因为它可以是多态的就认为它必须是多态的.通常，多态版本会更复杂.

There's no shame in defining a monomorphic function. Don't feel it has to be polymorphic just because it can be polymorphic. Often the polymorphic version is more complicated.

这是另一种方式:

lastNums :: (Integral a, Num a) => a -> a
lastNums = fromInteger . read . tail . show . toInteger

另一种方式:

lastNums :: (Integral a, Read a, Show a) => a -> a
lastNums = read . tail . show

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