如何将IO int转换为Haskell中的字符串？ [英] How to convert IO int to String in Haskell?

问题描述

我正在学习在Haskell中使用输入和输出。我试图生成一个随机数并将其输出到另一个文件。问题是随机数似乎返回IO int，我无法使用show将它转换为String。有人可以给我一个指针吗？

解决方案

如果您向我们展示您写的代码不是工作。

无论如何，你在一个 do 块并且写了类似的东西，是吗？

  main = do 
 ... 
 writeFilesome-file.txt（show generateRandomNumberSomehow）
 ... 
  

您应该这样做：

pre code $ main $

randomNumber< - generateRandomNumberSomehow
writeFilesome-file.txt（show randomNumber）
...

< - 运算符将右边的 IO Int 值的结果绑定到左边的 Int -valued变量。 （是的，您也可以使用它将 IO字符串值的结果绑定到 String -valued变量，等等）

这个语法只在 do 块内有效。请注意， do 块本身会产生一个IO值 - 您无法清除IO内存。

I'm learning to use input and output in Haskell. I'm trying to generate a random number and output it to another file. The problem is that the random number seems to be returning "IO int" that I can't covert it to String using "show". Could someone give me a pointer here??

解决方案

It's helpful if you show us the code you've written that isn't working.

Anyway, you are in a do block and have written something like this, yes?

main = do
    ...
    writeFile "some-file.txt" (show generateRandomNumberSomehow)
    ...

You should instead do something like this:

main = do
    ...
    randomNumber <- generateRandomNumberSomehow
    writeFile "some-file.txt" (show randomNumber)
    ...

The <- operator binds the result of the IO Int value on the right to the Int-valued variable on the left. (Yes, you can also use this to bind the result of an IO String value to a String-valued variable, etc.)

This syntax is only valid inside a do block. It's important to note that the do block will itself result in an IO value --- you can't launder away the IO-ness.

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