在Vim中注释c ++代码 [英] Commenting c++ codes in Vim

问题描述

我要映射\c来注释vim中的当前行（在第一个非空行之前添加'//'到行），并且\d删除行开头的'//' （取消注释）。

我在vimrc中添加了这两行：

  imap \c< Esc>：s / ^ / \ / \ /< CR> j $ a 
 
 imap \d< Esc>：s / ^ \ / \ / /< CR> j $ a 
  

行：

  for（int i = 0; i  
 > 
 
 
   
但我想转换to：
  // for（int i = 0; i   
我想要的一个原因是，当我想让vim缩进这一行，它会添加一个额外的标签并将其转换为：
  // for（int i = 0; i   
我该怎么办？我认为^指的是行的第一个非空白字符，但它不是真的。
 
 
 
 PS：我找到了这两个命令，但我不知道他们在干什么。你能请你向我解释（因为我需要记住他们，他们更容易记住，当我知道他们是什么）。如果它们的任何部分不必要请告诉我：
  imap \c< ESC>：s，^ \ \s * \）[^ / \t] \ @ =，\1 //，e< CR> j $ a 
 
 imap \d  ESC ，^ \（\s * \）// \s\ @ !, \1，e< CR> j $ a 
  
 
 
解决方案
关于您找到的这些命令：
 
 
 imap \c：s，^（\s *）[^＃\t] \ @ =，\1 //，ej $ a 
 
 
让我们从模式匹配开始，与：s / ^ / \ / \ /< CR> j $ a  。
 
 
 
 
 
 ：s，完全等价于： s /   - 你可以选择一个非标准的分隔符，使表达式更容易阅读，因此这里使用，而不是 表示更少的转义，并且注释 // 和分隔符之间没有混淆
 
 
  ^（\s *）匹配行的开头（与yours相同）和任何前导空格，捕获空格以便稍后重复使用
 
 
  [^＃\t] \ @ = 只需要一些空格，制表符或 在行结束之前，但不捕获它（即，该字符不会被替换）
 
 
 ，\1 //，用（\ c $ c）捕获的任何内容替换与前导空格（ \1  s *） group），其后是 //  
 
 
 
 
 < hr> 
 
 
例如：
  ^ void foo 
  
应该匹配： ^（\s *）[^＃ \t] \ @ =   - >  ^（）（v @ =）void foo（）; 前导空格和下一个表达式匹配但不消耗 v 从 void 。 
然后， \1 // 扩展为：
  ^ // 
  
，最终替换结果为：
  ^ // void foo（）;`
  
（我已经标记了 ^ 的开头，因为缩进是不明确的）
 
I want to map \c to comment the current line in vim (add '// ' to before the first non-blank to the line) and \d to delete the '// ' at the beginning of line(uncomment).

I've added these two lines in my vimrc:
imap \c <Esc>:s/^/\/\/ <CR>j$a

imap \d <Esc>:s/^\/\/ /<CR>j$a
but when I call this in this line:
    for (int i = 0; i < n; i++)
then it is converted to:
//    for (int i = 0; i < n; i++)
but I want to convert it to:
    // for (int i = 0; i < n; i++)
one of the reasons I want this is that when I want vim to indent this line, it will add an extra tab at the beginning of the line and convert it to:
    //    for (int i = 0; i < n; i++)
what should I do? I thought that ^ refers to first non-blank character of line but it wasn't true.

P.S: I've found these two commands, but I don't know what are they doing. could you please explain them to me (because I need to remember them and they're easier to remember when I know what are they). if any part of them is unnecessary please tell me:
imap \c <ESC>:s,^\(\s*\)[^/ \t]\@=,\1// ,e<CR>j$a

imap \d <ESC>:s,^\(\s*\)// \s\@!,\1,e<CR>j$a

解决方案
About those commands you found:

  
imap \c :s,^(\s*)[^# \t]\@=,\1// ,ej$a
Let's start with the pattern match, comparing it with your :s/^/\/\/ <CR>j$a.


:s, is exactly equivalent to :s/ - you can choose a nonstandard seperator where it makes the expression more readable, so here using , instead of / means less escaping, and no confusion between the comment // and the seperator
^(\s*) matches the beginning of the line (same as yours) and any leading whitespace, capturing the whitespace so we can reuse it later
[^# \t]\@= just requires some character other than space, tab or # before the line end, but doesn't capture it (ie, that character won't be replaced)
,\1// , replaces the matched string with the leading whitespace (\1 is whatever was captured by the (\s*) group) followed by //




For example: 
^    void foo();
should match like so: ^(\s*)[^# \t]\@= -> ^(    )(v@=)void foo();, ie, the first group matches the leading whitespace, and the next expression matches but doesn't consume the v from void.
Then, \1// expands to:
^    // 
and the final substitution result is:
^    // void foo();`
(I've marked the start of line with ^ since the indentation is ambiguous otherwise)

                        
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