在以下代码的第 4 行中，我在 do 块中的两个 IO 操作之间夹有一个表达式：
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))
我将 do 表示法理解为使用 >>= 或 >> 将单子操作链接在一起。但是当你有一个介于两者之间的表达式时，它是如何工作的呢？您不能仅使用 >> 将第 3-5 行粘合在一起。
我将从我非常相似的答案中抄袭here（尽管可能不是重复的，因为该问题没有明确处理let).
The Report给出了从 do 语法到 Haskell 内核的完整翻译；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}
所以你的代码脱糖如下：
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))
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