用c++计算文件中的代码行数，只用注释或空行跳过行

Question

我目前正在开发一个加载文件然后将行数读回给用户的程序。我只需要一些关于何时让程序计算一行的帮助。目前我有：

getline(file, line);
if(line.empty() || line.find("//") || line.find("**") || line.find("/*") || line.find("*/"))
{ skip line }

唯一的问题是如果程序仍有代码但有注释（以下任何示例），则程序不会计算行数：

code...  //comment
code... /* comment
/* comment */ code... 

任何提示或帮助？

Answer 1

完成您需要的算法可能如下：
（注：此代码未经测试）

std::vector<std::string> lines;
//collect all the lines in the vector

//iterate through the vector, removing all leading and trailing whitespace characters
for(auto& line : lines)
{
  std::string::size_type pos = line.find_last_not_of("\n\t ");
  if(pos != std::string::npos)
    line.erase(pos + 1)

  pos = line.find_first_not_of("\n\t ");
  if(pos != std::string::npos && pos != 0)
    line.erase(0, pos);
}

//remove all comment-only lines and blank lines
bool comment_block = false;
lines.erase(std::remove_if(lines.begin(), lines.end(), [&comment_block](const std::string& line){
  //if it's empty, remove it
  if(line.empty())
    return true;

  //if it starts with "//" it can't have code after it
  if(line.find("//") == 0)
    return true;

  //if it starts with "/*", it begins a comment block
  //(or if we're already in a comment block from earlier...)
  if(comment_block || line.find("/*") == 0)
  {
    auto pos = line.find("*/");
    if(pos == std::string::npos)
    {
      //if we can't find "*/" in this line, this is a comment block
      //so this line is all comments and it continues to the next line
      comment_block = true;
      return true;
    }
    //if we did find "*/" in this line, we are no longer in a comment block
    //(if we were already)
    comment_block = false;
    //Also, if "*/" is the last thing in this line, it's all comments
    return pos == line.length() - 2;
  }

  return false;
}), lines.end());