Question

在下面的代码第4行中，我有一个表达式夹在do块中的两个IO动作之间：

  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))

我理解使用＆gt;＆gt; =或＆gt;＆gt;链接monadic操作的符号。但是当你在两者之间有表达式时，它是如何工作的？你不能只使用＆gt;＆gt;。

将第3-5行粘在一起

Answer 1

我要从我非常相似的答案中here（尽管可能不重复，因为该问题并未明确处理let）。

Report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是：

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

所以你的代码就像这样去了：

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))

如何在Haskell中允许表达式阻塞

1 个答案: