等于（=）Vs左箭头（&lt ;-) haskell中的符号 [英] Equal (=) Vs left arrow (<-) symbols in haskell

问题描述

工作代码：

 导入系统
 main = do 
 [file1，file2]< -  getArgs 
  - 复制文件内容
 str<  -  readFile file1 
 writeFile file2 str 
  

崩溃的代码：

  import System 
 main = do 
 [file1， file2] = getArgs 
 str = readFile file1 
 writeFile file2 str 
  

当我尝试，它会抛出一个错误：

a.hs：6：18：解析输入错误'='

因此， < - 与 =


对于IO monad的具体情况，例如在 getArgs 示例中，一个粗略但有用的直觉可以做成如下：



• x < - action 运行 IO / code>，得到结果，并将它绑定到 x


• let x = action 定义 x 等价于 action ，但不会运行任何东西。稍后，您可以使用 y < - x ，这意味着 y < - action 。


$ b

来自允许闭包的命令式语言的程序员可以使用Javascript进行粗略的平行比较：

  var action = function（）{print（3）;返回5; } 
 
 //大致相当于x < -  action 
 print（'test 1'）
 var x = action（）//输出：3 
 / / x是5 
 
 //大致相当于让y =动作
 print（'test 2'）
 var y =动作//输出：nothing 
 / / y是函数
 
 //大致相当于z <-y 
 print（'test 3'）
 var z = y（）//输出：3 
 // z是5 
  

再次说明：此比较仅着重于IO。对于其他monads，您需要检查>> = 实际上是什么，并考虑解析 do 。




Working code:

import System
main = do
     [file1, file2] <- getArgs
     --copy file contents
     str <- readFile file1
     writeFile file2 str

Crashing code:

import System
main = do
       [file1, file2] = getArgs
       str = readFile file1
       writeFile file2 str

When I tried, it threw an error as:

a.hs:6:18: parse error on input '='

So, how different is <- from =?

解决方案

To understand the real difference, you have to understand monads, and the desugaring described by @rightfold in their answer.

For the specific case of the IO monad, as in your getArgs example, a rough but useful intuition can be made as follows:

• x <- action runs the IO action, gets its result, and binds it to x
• let x = action defines x to be equivalent to action, but does not run anything. Later on, you can use y <- x meaning y <- action.

Programmers coming from imperative languages which allow closures, may draw this rough parallel comparison with Javascript:

var action = function() { print(3); return 5; }

// roughly equivalent to x <- action
print('test 1')
var x = action()  // output:3
// x is 5

// roughly equivalent to let y = action
print('test 2')
var y = action    // output: nothing
// y is a function

// roughly equivalent to z <- y
print('test 3')
var z = y()       // output:3
// z is 5

Again: this comparison focuses on IO, only. For other monads, you need to check what >>= actually is, and think about the desugaring of do.

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