"< - "在符号中绑定 [英] "<-" bindings in do notation

问题描述

我很难把握这一点。在写符号时，以下两行如何不同？

  1。让x =表达式
 2. x < - 表达式
  

我看不到它。有时候一个人工作，一个人工作。但很少两个。 学习你一个haskell说，< - 将右侧绑定到左侧的符号。但是，与 let ？

< - 语句将从monad中提取值，并且 let
$ b

  import Data.Typeable 
 
 readInt :: String  - > IO Int 
 readInt s = do 
 putStrLn $为++ s ++输入值：
 readLn 
 
 main = do 
x< ;  -  readIntx
 let y = readInty
 putStrLn $x ::++ show（typeOf x）
 putStrLn $y ::++ show typeOf y）
  

运行时，程序会询问x的值，因为monadic action readIntx由< - 语句执行。它不会询问y的值，因为 readInty会被评估，但是不会执行产生的一元行为。

 
输入x的值：
 123 
x :: Int 
y :: IO Int 
 

$ b

既然 x :: Int ，你可以做普通的 Int



  putStrLn $x =++显示x 
 putStrLn $x * 2 =++ show（x * 2）
  

由于 y :: IO Int ，你不能假装它是一个普通的 Int 。

  putStrLn $y =++ show y  - 错误
 putStrLn $y * 2 =++ show（y * 2） - 错误
  




I have a hard time grasping this. When writing in do notation, how are the following two lines different?

1. let x = expression
2. x <- expression

I can't see it. Sometimes one works, some times the other. But rarely both. "Learn you a haskell" says that <- binds the right side to the symbol on the left. But how is that different from simply defining x with let?

解决方案

The <- statement will extract the value from a monad, and the let statement will not.

import Data.Typeable

readInt :: String -> IO Int
readInt s = do
  putStrLn $ "Enter value for " ++ s ++ ": "
  readLn

main = do
  x <- readInt "x"
  let y = readInt "y"
  putStrLn $ "x :: " ++ show (typeOf x)
  putStrLn $ "y :: " ++ show (typeOf y)

When run, the program will ask for the value of x, because the monadic action readInt "x" is executed by the <- statement. It will not ask for the value of y, because readInt "y" is evaluated but the resulting monadic action is not executed.

Enter value for x: 
123
x :: Int
y :: IO Int

Since x :: Int, you can do normal Int things with it.

putStrLn $ "x = " ++ show x
putStrLn $ "x * 2 = " ++ show (x * 2)

Since y :: IO Int, you can't pretend that it's a regular Int.

putStrLn $ "y = " ++ show y -- ERROR
putStrLn $ "y * 2 = " ++ show (y * 2) -- ERROR

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