Haskell中函数的序列化 [英] Serialization of functions in Haskell

问题描述

有没有一种方法可以在Haskell中序列化（读取/显示）函数？例如，给出：

 ：t（+1）
（+1）:: Num a => a  - > a 
  

我希望能有这样的东西：

 读取（+1）:: Num a => a  - > a 
  

不幸的是，这会抛出一个错误：

 无法推断（读取（a  - > a））由上下文（数字a）
使用读取
引起类型签名：Num a => a  - > a 
 at< interactive> 1：1-30 
可能的修正：
 add（读取（a  - > a））到
的上下文中表达式类型签名：Num a => a  - > a 
或者添加一个实例声明（Read（a  - > a））
在表达式中：read（+1）:: Num a => a  - > a 
在'it'的等式中：it = read（+1）:: Num a => a  - > a 
  

解决方案

您可以使用 plugins 包在运行时读取代码。正如augustss所说，显示是不可能的。

一个可以如何使用的例子：

  import System.Eval.Haskell 
 
 main = do 
 mf<  -  eval（+1）:: Int  - > Int[] 
案例mf of 
只需f  - > print $（f :: Int  - > Int）0 
 _  - > putStrLn由于某些原因无法评估。:(
  

Is there a way to serialize (read/show) functions in Haskell?

For example given that:

:t (+1) 
(+1) :: Num a => a -> a

I wish to be able to have something like:

read "(+1)" :: Num a => a -> a

Unfortunately this throws an error:

Could not deduce (Read (a -> a)) arising from a use of `read'
from the context (Num a)
  bound by an expression type signature: Num a => a -> a
  at <interactive>:1:1-30
Possible fix:
  add (Read (a -> a)) to the context of
    an expression type signature: Num a => a -> a
  or add an instance declaration for (Read (a -> a))
In the expression: read "(+1)" :: Num a => a -> a
In an equation for `it': it = read "(+1)" :: Num a => a -> a

解决方案

You could use something like the plugins package to read code at runtime. Showing is, as augustss says, impossible though.

An example of how it could be used:

import System.Eval.Haskell

main = do
  mf <- eval "(+1) :: Int -> Int" []
  case mf of
    Just f -> print $ (f :: Int -> Int) 0
    _      -> putStrLn "Couldn't eval for some reason. :("

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