Haskell语法错误where语句 [英] Haskell syntax error for where statement

问题描述

Vec2.hs： 33:27：解析输入错误'='

我在这里写的代码如下。错误指向第二项 vec2Normalize iLength = ... 我没有看到语法错误

   - 获得v的逆长度并将其乘以它
  - 得到v的归一化形式
 vec2Normalize :: Vec2  - > Vec2 
 vec2Normalize v @（x，y）=（x * iLength，y * iLength）
其中length = vec2Length v 
 iLength = if长度== 0 then 1 else（1 / length ）
  

解决方案

由于您不提供完整的代码，但是这个错误可能表明你的行 iLength = ... 没有正确缩进;实际上， iLength 开始于前一行的 length = 的 right 。

您的原始文件是否使用制表符而不是空格进行缩进？如果是这样，请注意，Haskell总是将选项卡解释为跨越8列。因此，例如，

 < TAB>其中，长度= ... 
< TAB>< TAB> < SPACE>< SPACE> iLength = ... 
  

会被解释为

  where length = ... 
 iLength = ... 
  

因此会导致错误，即使您的编辑器在使用4列选项卡时可能会显示正确对齐的行。

I'm writing some Haskell code to learn the language, and I've run into the syntax error:

Vec2.hs:33:27: parse error on input '='

The code I've written here is below. The error is pointing at the 2nd term in vec2Normalize iLength = ... I don't see the syntax error

-- Get the inverse length of v and multiply the components by it
-- Resulting in the normalized form of v
vec2Normalize :: Vec2 -> Vec2
vec2Normalize v@(x,y) = (x * iLength, y * iLength)
    where length = vec2Length v
          iLength = if length == 0 then 1 else (1 / length)

解决方案

Some guessing involved since you don’t provide the complete code, but this error could indicate that your line iLength = ... is not properly indented; actually, that the iLength starts to the right of the length = on the line before.

Does your original file use tabs instead of spaces for indentation? If so, be aware that Haskell always interprets a tab as spanning 8 columns. So, e.g.,

<TAB>where length = ...
<TAB><TAB><SPACE><SPACE>iLength = ...

would be interpreted as

        where length = ...
                  iLength = ...

thus causing the error, even though your editor might show the lines properly aligned if it uses 4-column tabs.

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