Question

我需要打上原木和I've 好的工作reg,但现在我需要从编织档案中 set,这里是问题。

int logParser()
{
  std::string bd_regex; // this reads from config in other part of program
  boost::regex parsReg;
  //("(C:.tmp.bd.*?)+(([a-zA-Z0-9_]+\.)+[a-zA-Z]{2,4})+(.+[a-zA-Z0-9_])");
  try
  {
    parsReg.assign(bd_regex, boost::regex_constants::icase);  
  }
  catch (boost::regex_error& e)
  {
    cout << bd_regex << " is not a valid regular expression: ""
         << e.what() << """ << endl;
  }

  cout << parsReg << endl;
  // here it looks exactly like:
  // "("(C:.tmp.bd.*?)+(([a-zA-Z0-9_]+\.)+[a-zA-Z]{2,4})+(.+[a-zA-Z0-9_])");"

  int count=0;
  ifstream in;

  in.open(bd_log_path.c_str());

  while (!in.eof()) 
  {
    in.getline(buf, BUFSIZE-1);
    std::string s = buf;
    boost::smatch m;

    if (boost::regex_search(s, m, parsReg)) // it doesn t obey this "if"
    {
      std::string name, diagnosis;
      name.assign(m[2]);
      diagnosis.assign(m[4]);

      strcpy(bd_scan_results[count].file_name, name.c_str());
      strcpy(bd_scan_results[count].out,  diagnosis.c_str());
      strcat(bd_scan_results[count].out,  " ");

      count++;
      } 
    }
  return count;
}

我确实很想知道,为什么在我试图将其从混杂变量中确定下来时,同样的reg勤工作。

Any help will be appreciated (:

Answer 1

On your direct question: Try storing the regex without escapes in the config file

(C:.tmp.bd.*?)+(([a-zA-Z0-9_]+.)+[a-zA-Z]{2,4})+(.+[a-zA-Z0-9_])

_{Besides,我必须说,它同你一样,想在这里对lash进行对应:}。

C:.tmp.bd.

书写:

C:\tmp\bd\

缩略语

"C:\\tmp\\bd\\"

Answer 2

@sehe给出正确的答案。

If this line of code were parsed by the c++ parser,
str = "(C:.tmp.bd.*?)+(([a-zA-Z0-9_]+\.)+[a-zA-Z]{2,4})+(.+[a-zA-Z0-9_])";

it would unescape the escape character \ into just an escape: , then
asign it to variable str . Inside of the variable str , it now looks like this:
(C:.tmp.bd.*?)+(([a-zA-Z0-9_]+.)+[a-zA-Z]{2,4})+(.+[a-zA-Z0-9_])

But, you are reading this text from a file, there is no parsing in a language sense.
You are asigning to str , a raw line of text. A line that is not pre-processed by the c++ parser.

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