Question

我已经学习了Haskell一段时间了，但IO monad却吓坏了我。我有一个代码

main = do
    putStrLn "First computation starts"
    let firstResult = foo -- foo is a pure function
    putStrLn "Second computation starts"
    let secondResult = bar foo -- bar is too pure function
    writeFile secondResult

然后看"First computation starts" "Second computation starts"，然后程序即可。

我知道有laisy计算，真正的计算在writeFile执行时开始。我尝试添加严格性

    main = do
    putStrLn "First computation starts"
    let !firstResult = foo -- foo is a pure function
    putStrLn "Second computation starts"
    let !secondResult = bar foo -- bar is too pure function
    writeFile secondResult

什么都没发生

那么，也许让表达式只是同义词并由编译器交换？我尝试将函数转换为IO

main = do
putStrLn "First computation starts"
!firstResult <- return (foo) -- foo is a pure function
putStrLn "Second computation starts"
!secondResult <- return (bar foo) -- bar is too pure function
writeFile secondResult

尽管如此，结果与上述相同，对我来说还不清楚。

Answer 1

如果您想确定某些内容已被评估，deepseq是您的朋友。无论NFData是什么，您都必须实施foo，但这通常很容易实现。您的main将成为

main = do
  putStrLn "First computation starts"
  firstResult <- return $!! foo
  putStrLn "Second computation starts"
  secondResult <- return $!! bar foo
  writeFile secondResult

IO计算顺序

1 个答案: