为什么不能在Haskell中添加一个Int和一个浮点数 [英] why cant an Int and a floating point number be added in haskell

问题描述

 （长度[1,2,3,4]）+ 3.2 
  

尽管如此： -

  2 + 3.3 
  

我知道在第一种情况下，结果是一个Int +浮点数，但在第二种情况下也不一样，或者Haskell在第二种情况下自动推断出类型为： - Num + Num，而在第一种情况下不会这样做？

解决方案

Haskell 从不为您进行隐式类型转换。 + 只能处理两个相同类型的数字，并且也会给出该类型作为结果。正如您在（长度[1,2,3,4]）+ 3.2 中看到的那样， + 的任何其他用法都是错误的c $ c> example。

但是，数字文字在Haskell中被重载。 2 可以是任何数字类型， 3.3 可以是任何小数类型。所以当Haskell看到表达式 2 + 3.3 时，它可以尝试找到一个既是数字也是小数的类型，并将这两个数字视为该类型，

更准确地说， + 的类型为 Num a => ; a - > a - >一个。 2 本身是类型 Num a =>一个和 3.3 本身是类型小数a =>一个。将这3种类型放在一起，在 2 + 3.3 这两个表达式中可以给出类型 Fractional a =>一个，因为所有小数类型也是 Num 类型，这也满足类型 + 。 （如果你在GHCi中输入这个表达式， a 被填充为 Double ，因为GHC必须将类型默认为 为了评估它）

在表达式中（长度[1,2,3,4]） + 3.2 ， 3.2 仍然被重载（并且隔离的类型为 Fractional a => a ）。但长度[1,2,3,4] 的类型为 Int 。由于一方是固定的具体类型，满足 + 类型的唯一方法是填写 a 在 Int 的其他类型上，但违反了小数约束;没有办法让 3.2 成为 Int 。所以这个表达式不是很好的类型。然而，任何 Integral 类型（其中 Int 是一个）可以通过应用 fromIntegral 来转换为任何 Num code>（这实际上是如何将像 2 这样的整数文字视为任何数字类型）。因此，（fromIntegral $ length [1,2,3,4]）+ 3.2 将可用。

why wont this work :-

(length [1,2,3,4]) + 3.2

while this works:-

2+3.3

I understand that in the first case the result is an Int+Float but is that not the same in the second case too, or does Haskell automatically infer the type in the second case to be :- Num+Num whereas it does not do that in the first case?

解决方案

Haskell never does implicit type conversion for you. + only ever works on two numbers of the same type, and gives you that type as the result as well. Any other usage of + is an error, as you saw with your (length [1,2,3,4]) + 3.2 example.

However, numeric literals are overloaded in Haskell. 2 could be any numeric type, and 3.3 could be any fractional type. So when Haskell sees the expression 2 + 3.3 it can try to find a type which is both "numeric" and "fractional", and treat both numbers as that type so that the addition will work.

Speaking more precisely, + has the type Num a => a -> a -> a. 2 on its own is of type Num a => a and 3.3 on its own is of type Fractional a => a. Putting those 3 types together, the in the expression 2 + 3.3 both numbers can be given the type Fractional a => a, because all Fractional types are also Num types, and this then also satisfies the type of +. (If you type this expression into GHCi the a gets filled in as Double, because GHC has to default the type to something in order to evaluate it)

In the expression (length [1,2,3,4]) + 3.2, the 3.2 is still overloaded (and in isolation would have type Fractional a => a). But length [1,2,3,4] has type Int. Since one side is a fixed concrete type, the only way to satisfy the type for + would be to fill in the a on the other type with Int, but that violates the Fractional constraint; there's no way for 3.2 to be an Int. So this expression is not well-typed.

However, any Integral type (of which Int is one) can be converted to any Num type by applying fromIntegral (this is actually how the integer literals like 2 can be treated as any numeric type). So (fromIntegral $ length [1,2,3,4]) + 3.2 will work.

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