问题描述
Haskell 中的类型推断有一点学习曲线(至少可以这么说！).开始学习它的一个好方法是使用简单的示例.因此，以下是类型推断的hello world".
Type inference in Haskell has a bit of a learning curve (to say the least!). A good way to start learning it is with simple examples. So, the following is a bit of a "hello world" for type inference.
考虑以下示例:
Prelude> :t 3
3 :: (Num t) => t
Prelude> let x = 3
Prelude> :t x
x :: Integer
问题是这样的:为什么 3 和 x 有不同的类型?
链接摘要:
阅读以下答案以了解完整故事；这里只是一个链接摘要:
Read the answers below for the full story; here's just a link summary:
- GHC 类型默认值:Haskell 报告部分4.3.4
- GHCi 的扩展类型默认值:使用 GHCi 部分2.4.5
- 单态限制:Haskell维基
推荐答案
这里还有另一个因素，在 acfoltzer 包含的一些链接中提到，但在此处可能值得明确说明.您遇到了单态限制的影响.当你说
There's another factor here, mentioned in some of the links which acfoltzer includes, but it might be worth making explicit here. You're encountering the effect of the monomorphism restriction. When you say
let x = 5
您对变量进行了顶级定义.MR 坚持认为，当这些定义没有类型签名时，应该通过为未解析的类型变量选择(希望)合适的默认实例来专门化为单态值.相比之下，当您使用 :t 请求推断类型时，不会强加此类限制或默认设置.所以
you make a top-level definition of a variable. The MR insists that such definitions, when otherwise unaccompanied by a type signature, should be specialized to a monomorphic value by choosing (hopefully) suitable default instances for the unresolved type variables. By contrast, when you use :t to ask for an inferred type, no such restriction or defaulting is imposed. So
> :t 3
3 :: (Num t) => t
因为 3 确实是重载的:它被任何数字类型承认.默认规则选择Integer作为默认的数字类型，所以
because 3 is indeed overloaded: it is admitted by any numeric type. The defaulting rules choose Integer as the default numeric type, so
> let x = 3
> :t x
x :: Integer
但是现在让我们关闭 MR.
But now let's switch off the MR.
> :set -XNoMonomorphismRestriction
> let y = 3
> :t y
y :: (Num t) => t
如果没有 MR，定义就尽可能多态，就像 3 一样重载.只是检查...
Without the MR, the definition is just as polymorphic as it can be, just as overloaded as 3. Just checking...
> :t y * (2.5 :: Float)
y * (2.5 :: Float) :: Float
> :t y * (3 :: Int)
y * (3 :: Int) :: Int
请注意，根据相关 Num 实例提供的 fromInteger 方法，多态 y = 3 在这些用途中具有不同的专门化.也就是说，y 与3 的特定表示无关，而是与3 的表示构造相关联的方案.天真地编译，这是缓慢的秘诀，有些人将其作为 MR 的动机.
Note that the polymorphic y = 3 is being differently specialized in these uses, according to the fromInteger method supplied with the relevant Num instance. That is, y is not associated with a particular representation of 3, but rather a scheme for constructing representations of 3. Naïvely compiled, that's a recipe for slow, which some people cite as a motivation for the MR.
我(在本地假装是)在关于单态限制是较小还是较大的邪恶的辩论中保持中立.我总是为顶级定义编写类型签名，所以我想要实现的目标没有歧义，MR 不在重点.
I'm (locally pretending to be) neutral on the debate about whether the monomorphism restriction is a lesser or greater evil. I always write type signatures for top-level definitions, so there is no ambiguity about what I'm trying to achieve and the MR is beside the point.
在尝试了解类型系统如何工作时，将类型推断的各个方面分开是非常有用的
When trying to learn how the type system works, it's really useful to separate the aspects of type inference which
‘遵循计划’，将多态定义专门用于特定用例:约束解决的一个相当强大的问题，需要通过反向链进行基本的统一和实例解析；和
‘follow the plan’, specializing polymorphic definitions to particular use cases: a fairly robust matter of constraint-solving, requiring basic unification and instance resolution by backchaining; and
‘猜计划’，概括类型以将多态类型方案分配给没有类型签名的定义:这非常脆弱，而且越是越过基本的 Hindley-Milner 学科，类型类，有了更高等级的多态性，有了 GADT，事情就变得奇怪了.
‘guess the plan’, generalizing types to assign a polymorphic type scheme to a definition with no type signature: that's quite fragile, and the more you move past the basic Hindley-Milner discipline, with type classes, with higher-rank polymorphism, with GADTs, the stranger things become.
了解第一个是如何工作的，并理解为什么第二个是困难的，这很好.类型推断中的许多怪异之处都与第二个相关，并且与单态限制之类的启发式方法试图在面对歧义时提供有用的默认行为.
It's good to learn how the first works, and to understand why the second is difficult. Much of the weirdness in type inference is associated with the second, and with heuristics like the monomorphism restriction trying to deliver useful default behaviour in the face of ambiguity.
这篇关于为什么 3 和 x(被赋值为 3)在 Haskell 中有不同的推断类型?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，WP2