您构造一个putStrLn "1"作为最终结果，并由此生成。注意IO a * 不是 * 一个 * 发生过 * 的动作，或者总是会发生，你应该把它看作是动作的某种委托，所以是一个配方，你可以把合并配方和**(>>) :: Monad m => m a -> m b -> m b**结合起来，创建一个 * 新的 * 配方，它首先运行第一个操作数，然后是正确的操作数。
所以在这种情况下，我们可以用以下方法修复它：

foldLeft (\prev x -> prev >> putStrLn x) (pure ()) ("2":."1":.Nil)

所以最终结果将是putStrLn "2" >> putStrLn "1"，它是IO ()，当调用时将首先打印"2"，然后打印"1"。

因为Haskell是引用透明的，所以你总是可以通过手动计算表达式来回答这样的问题，即内联函数调用和beta缩减，直到很明显发生了什么：

foldLeft (\_ x -> putStrLn x) (pure ()) ("2":."1":.Nil)
  = {- (recursive) definition of `foldLeft` for Cons case -}
    foldLeft (\_ x -> putStrLn x)
       ((\_ x -> putStrLn x) (pure ()) "2")
       ("1":.Nil)
  = {- beta reduction -}
    foldLeft (\_ x -> putStrLn x)
       (putStrLn "2")
       ("1":.Nil)
  = {- definition of `foldLeft` for Cons case -}
    foldLeft (\_ x -> putStrLn x)
       ((\_ x -> putStrLn x) (putStrLn "2") "1")
       Nil
  = {- beta reduction -}
    foldLeft (\_ x -> putStrLn x)
       (putStrLn "1")  -- woops, the "2" has vanished in the `_` argument
       Nil
  = {- definition of `foldLeft` for Nil case -}
    putStrLn "1"

所以这就是你得到的：一张"1"的照片奥托赫，根据威廉货车昂塞姆的建议，

foldLeft (\prev x -> prev >> putStrLn x) (pure ()) ("2":."1":.Nil)
  = {- definition of `foldLeft` for Cons case -}
    foldLeft (\prev x -> prev >> putStrLn x)
       ((\prev x -> prev >> putStrLn x) (pure ()) "2")
       ("1":.Nil)
  = {- beta reduction -}
    foldLeft (\prev x -> prev >> putStrLn x)
       (pure () >> putStrLn "2")
       ("1":.Nil)
  = {- definition of `foldLeft` for Cons case -}
    foldLeft (\prev x -> prev >> putStrLn x)
       ((\prev x -> prev >> putStrLn x) (pure () >> putStrLn "2") "1")
       Nil
  = {- beta reduction -}
    foldLeft (\prev x -> prev >> putStrLn x)
       ((pure () >> putStrLn "2") >> putStrLn "1")
       Nil
  = {- definition of `foldLeft` for Nil case -}
    pure () >> putStrLn "2" >> putStrLn "1"