有没有像使用CString一样链接函数的方法？ [英] Is there a way to chain functions like withCString?

问题描述

有没有办法像 with CString 那样链接函数？我的意思是任何
函数看起来像 f :: Foo - > （CFoo→IOa）→> IO a 。

例如，假设有一个函数 cFunc :: CString - > CFoo - > CBar - > IO（）

Usualy，我会这样做：

  haskellFunc string foo bar = 
 withCString string $ \ cString  - > 
 withCFoo foo $ \ cFoo  - > 
 withCBar bar $ \ cBar  - > 
 cFunc cString cFoo cBar 
  

但是我想要这样做：

  haskellFunc =（withCString |。| withCFoo |。| withCBar）cFunc 
  

与一些适当的组合运算符 |。| 。

我正在写一个带有很多C绑定的库，而且这个样板经常是
。我做错了什么？

package / mtl-2.2.1 / docs / Control-Monad-Cont.html> Cont inuation applicative 来编写这些 a - > （b→IOc）→> IO c 函数：

  import Control.Monad.Cont 
 
 haskellFunc :: String  - > Foo  - >酒吧 - > IO（）
 haskellFunc字符串foo bar =翻转runCont id $ 
 cFunc< $> 
 cont（withCString string）< *> 
 cont（with CFoo foo）*< 
 cont（withCBar bar）
  

或者添加一些额外的语法：

  haskellFunc':: String  - > Foo  - >酒吧 - > IO（）
 haskellFunc'string foo bar = flip runCont id $ 
 cFunc<<>>> withCString string<<>> withCFoo foo<<>> withCBar bar 
 where 
 f<< $>> x = f< $> cont x 
 f *<>> x = f *续x 
  

Is there a way to chain functions like withCString? By that I mean any function that looks something like f :: Foo -> (CFoo -> IO a) -> IO a.

For example, lets say there is a function cFunc :: CString -> CFoo -> CBar -> IO ()

Usualy, I would do something like:

haskellFunc string foo bar =
  withCString string $ \ cString ->
    withCFoo foo $ \ cFoo ->
      withCBar bar $ \ cBar ->
        cFunc cString cFoo cBar

But i would like to do something like:

haskellFunc = (withCString |.| withCFoo |.| withCBar) cFunc

with some appropriate composition operator |.|.

I'm writing library with a lot of C bindings, and this boilerplate comes often. Am I doing something wrong?

解决方案

You can use the Continuation applicative for composing these a -> (b -> IO c) -> IO c functions:

import Control.Monad.Cont

haskellFunc :: String -> Foo -> Bar -> IO ()
haskellFunc string foo bar = flip runCont id $ 
    cFunc <$> 
      cont (withCString string) <*> 
      cont (withCFoo foo) <*> 
      cont (withCBar bar)

Or with a bit of extra syntax:

haskellFunc' :: String -> Foo -> Bar -> IO ()
haskellFunc' string foo bar = flip runCont id $
    cFunc <<$>> withCString string <<*>> withCFoo foo <<*>> withCBar bar
  where
    f <<$>> x = f <$> cont x
    f <<*>> x = f <*> cont x

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