在没有上下文的情况下，Haskell中的return 5的类型是什么? [英] What is the type of return 5 in Haskell when no context is given?

问题描述

在此问题中，OP询问表达式的类型 return 5 是，并且已经在该问题中给出了答案:它是通用类型，可以通过键入来验证

In this question the OP asks what the type of the expression return 5 is and the answer has already been given in that question: it is a generic type, as can be verified by typing

:t return 5

在Haskell解释器中:

in the Haskell interpreter:

return 5 :: (Num a, Monad m) => m a

return 的具体实现取决于它出现的上下文:类型推断会将 m 限制为特定的monad，例如 Maybe ， [] ， IO 等.

The specific implementation of return is determined by the context in which it appears: type inference will restrict m to a specific monad such as Maybe, [], IO, and so on.

我还可以通过指定类型来强制解释器选择特定的monad.

I can also force the interpreter to pick a specific monad by specifying the type, e.g.

Prelude> return 5 :: Maybe Int
Just 5
Prelude> return 5 :: [Int]
[5]

以此类推.

现在，如果我输入表达式 return 5 而未指定类型，则会得到:

Now if I type the expression return 5 without specifying a type, I get:

Prelude> return 5
5

这让我感到很惊讶:我希望解释器告诉我它不能选择 return 的适当实现，因为它不能推断要使用的单子类型.

which is quite surprising to me: I would rather have expected the interpreter to tell me that it cannot pick an appropriate implementation of return because it cannot infer the monadic type to be used.

所以我的问题是:Haskell在这里使用了什么特定的monad?而这个单子是根据什么标准选择的呢?

So my question is: what specific monad has Haskell used here? And based on what criteria was this monad chosen?

编辑

感谢您的回答！实际上，如果我尝试编译该程序:

Thanks for the answer! In fact, if I try to compile this program:

module Main
where

a = return 5

main :: IO ()
main = putStrLn "This program should not compile"

我得到一个错误:

No instance for (Monad m0) arising from a use of `return'
The type variable `m0' is ambiguous
Relevant bindings include
  a :: m0 Integer (bound at invalid_return.hs:4:1)
Note: there are several potential instances:
  instance Monad ((->) r) -- Defined in `GHC.Base'
  instance Monad IO -- Defined in `GHC.Base'
  instance Monad [] -- Defined in `GHC.Base'
  ...plus one other
In the expression: return 5
In an equation for `a': a = return 5

因此，仅出于乔恩(Jon)解释的原因，它只能在GHCi中工作.

So it only works in GHCi for the reasons explained by Jon.

推荐答案

monad是 IO .这是GHCi行为的一个小怪癖.它尝试使用 IO a 统一输入的类型；如果成功，它将运行该 IO 操作并尝试 show 结果.如果您给它一个 IO 动作以外的东西，它只会尝试显示值.

The monad is IO. This is a minor quirk of the behaviour of GHCi. It tries to unify the type of your input with IO a; if it succeeds, it runs that IO action and tries to show the result. If you give it something other than an IO action, it simply tries to show the value.

出于相同的原因，它们产生相同的输出:

It’s for the same reason that these produce the same output:

Prelude> "hello"
"hello"
Prelude> print "hello"
"hello"

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