Convert string with explicit escape sequence into relative character

Question

I need a function to convert "explicit" escape sequences into the relative non-printable character. Es:

char str[] = "\\n";
cout << "Line1" << convert_esc(str) << "Line2" << endl:

would give this output:

Line1

Line2

Is there any function that does this?

Answer 1

I think that you must write such function yourself since escape characters is a compile-time feature, i.e. when you write "\n" the compiler would replace the \n sequence with the eol character. The resulting string is of length 1 (excluding the terminating zero character).

In your case a string "\\n" is of length 2 (again excluding terminating zero) and contains \ and n.

You need to scan your string and when encountering \ check the following char. if it is one of the legal escapes, you should replace both of them with the corresponding character, otherwise skip or leave them both as is.

( http://ideone.com/BvcDE ):

string unescape(const string& s)
{
  string res;
  string::const_iterator it = s.begin();
  while (it != s.end())
  {
    char c = *it++;
    if (c == '\\' && it != s.end())
    {
      switch (*it++) {
      case '\\': c = '\\'; break;
      case 'n': c = '\n'; break;
      case 't': c = '\t'; break;
      // all other escapes
      default: 
        // invalid escape sequence - skip it. alternatively you can copy it as is, throw an exception...
        continue;
      }
    }
    res += c;
  }

  return res;
}

Answer 2

You can do that fairly easy, using the boost string algorithm library. For example:

#include <string>
#include <iostream>
#include <boost/algorithm/string.hpp>

void escape(std::string& str)
{
  boost::replace_all(str, "\\\\", "\\");
  boost::replace_all(str, "\\t",  "\t");
  boost::replace_all(str, "\\n",  "\n");
  // ... add others here ...
}

int main()
{
  std::string str = "This\\tis\\n \\\\a test\\n123";

  std::cout << str << std::endl << std::endl;
  escape(str);
  std::cout << str << std::endl;

  return 0;
}

This is surely not the most efficient way to do this (because it iterates the string multiple times), but it is compact and easy to understand.

Update: As ybungalobill has pointed out, this implementation will be wrong, whenever a replacement string produces a character sequence, that a later replacement is searching for or when a replacement removes/modifies a character sequence, that should have been replaced.

An example for the first case is "\\\\n" -> "\\n" -> "\n". When you put the "\\\\" -> "\\" replacement last (which seems to be the solution at a first glance), you get an example for the latter case "\\\\n" -> "\\\n". Obviously there is no simple solution to this problem, which makes this technique only feasible for very simple escape sequences.

If you need a generic (and more efficient) solution, you should implement a state machine that iterates the string, as proposed by davka.

Answer 3

I'm sure that there is, written by someone, but it's so trivial that I doubt it's been specifically published anywhere.

Just recreate it yourself from the various "find"/"replace"-esque algorithms in the standard library.

Answer 4

Here's a cute way to do it on Unixy platforms.

It calls the operating system's echo command to make the conversion.

string convert_escapes( string input )
   {
   string buffer(input.size()+1,0);
   string cmd = "/usr/bin/env echo -ne \""+input+"\"";
   FILE * f = popen(cmd.c_str(),"r"); assert(f);
   buffer.resize(fread(&buffer[0],1,buffer.size()-1,f));
   fclose(f);
   return buffer;
   }

Answer 5

Have you considered using printf? (or one of its relatives)