运行简单的Haskell程序时解析错误 [英] Parse error while running a simple Haskell program

问题描述

我刚开始学习Haskell。我开始写一个简单的函数，它接受两个值n和x，然后显示用它计算出的两个整数值。

  test nx = do 
 cell1 = round（n * n * x）
 cell2 = n * n  -  cell1 
 print cell1 
 print cell2 
  

但它不运行并且一直给我输入错误的解析错误。发生了什么事？ 解决方案

你可能想要的是在你的 do 块中有一个 let 语句

  test nx = do 
 let cell1 = round（n * n * x）
 cell2 = n * n  -  cell1 
 print cell1 
 print cell2 
  

这里的区别是你不能直接在 do 块，因为所有 do 块都会被调用为>>> = 和>> 。 let 语句允许您定义一个本地值，就像您可以在函数定义中一样定义，如

  fx = 
 let y = 2 * x 
z = y * y * y 
 in z + z + y 
  

函数的解析方式就像

  test nx = 
让cell1 = round（n * n * x）
 cell2 = n * n  -  cell1 
 in（print cell1>> print cell2）
  

其中>> 只是将两个monadic动作链接在一起。请注意，这不是真的如何解除，我选择了一个在这种情况下等价的表示，但它不完全是编译器实际生成的。

I have just started learning Haskell. I'm beginning by writing a simple function that takes two values n and x, and then displays two integer values computed with it.

test n x = do
    cell1 = round(n*n*x)
    cell2 = n*n - cell1
    print cell1
    print cell2

But it doesn't run and keeps giving me a Parse error on input `=' error. What is happening?

解决方案

You've hit your first trouble with Monads. What you probably want here is a let statement inside your do block

test n x = do
    let cell1 = round (n * n * x)
        cell2 = n * n - cell1
    print cell1
    print cell2

The difference here is that you can't assign directly inside of a do block, since all do blocks desugar to calls to >>= and >>. The let statement allows you to define a local value like you can inside a function definition like

f x =
    let y = 2 * x
        z = y * y * y
    in z + z + y

The way your function would desugar would be like

test n x =
    let cell1 = round (n * n * x)
        cell2 = n * n - cell1
    in (print cell1 >> print cell2)

Where >> just chains two monadic actions together. Note that this is not really how it desugars, I chose a representation that is equivalent in this case but it is not exactly what the compiler would actually generate.

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