Haskell语法错误where语句

本文介绍了Haskell语法错误where语句的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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.