在Haskell do块中如何允许表达式

Question

在下面的代码第4行中，有一个表达式夹在do块中的两个IO操作之间：

  1 doubleX :: (Show x, Num x) => x -> IO ()                                                                                                                                                                                          
  2 doubleX x = do                                                                                                                                                                                                                    
  3   putStrLn ("I will now double " ++ (show x))                                                                                                                                                                                     
  4   let double = x * 2                                                                                                                                                                                                              
  5   putStrLn ("The result is " ++ (show double))

我理解do表示法是使用>>=或>>将一元操作链接在一起。但是，当你在中间有一个表达式时，它是如何工作的呢？您不能仅仅使用>>将3-5行粘合在一起。

Answer 1

我将从我非常相似的答案这里 (虽然可能不是一个重复的问题，因为这个问题没有明确地处理let)。

报告提供了从do语法到内核Haskell的完整翻译；与您的问题相关的部分如下：

do {e} =e do {e;stmts} =e >> } do {stmts} do {let decls；stmts} = let decls in do {stmts}

所以你的代码像这样：

doubleX x = do                                                                                                                                                                                                                    
  putStrLn ("I will now double " ++ (show x))                                                                                                                                                                                     
  let double = x * 2                                                                                                                                                                                                              
  putStrLn ("The result is " ++ (show double))

==> do {e;stmts} rule

doubleX x =
  putStrLn ("I will now double " ++ (show x)) >> do
  let double = x * 2                                                                                                                                                                                                              
  putStrLn ("The result is " ++ (show double))

==> do {let decls; stmts} rule

doubleX x =
  putStrLn ("I will now double " ++ (show x)) >>
  let double = x * 2 in do
  putStrLn ("The result is " ++ (show double))

==> do {e} rule

doubleX x =
  putStrLn ("I will now double " ++ (show x)) >>
  let double = x * 2 in
  putStrLn ("The result is " ++ (show double))