Passing on a char* which contains a path

Question

I'm trying to pass through several functions a string with a path but the every '\\\\' I put in the path becomes a single '\\' in the inner function and I can't use it this way.
Is there a way to preserve the "\\\\" when entering a new function?
I'm using C++ on windows.
thanks :)

Answer 1

Be prepared for some obfuscated answer.

The \\ is the escape character (you have probably already encountered the \\n escape sequence for example), and \\\\ is the escape sequence that represents a single \\ character (in a sense, it can be understood as an escape of the escape character). If you really want to have \\\\ in your string, you'll have to use \\\\\\\\ :

std::cout << "\\\\something\\" << std::endl; /* prints "\\something\" */

Just to provide another example, suppose you'd like to have some " in a string. Writing :

const char *str = "Hello "World"";

will obviously not compile, and you will have to escape the " with a \\ :

const char *str = "Hello \"World\"";

Answer 2

In C++0x you will have a raw string literal:

R"(anything can appear here, even " or \\ )"

Where everything between "( and )" is part of the string -- no escaping necessary. In the current standard you can't achieve what you want.