I'm using VS2012/C++, I need to convert a std::string to char * and I can't find any material online giving any guidance on how to go about doing it.
Any code examples and advice will be greatly appreciated.
1 Answer
Use
std::string bla("bla");
char* blaptr = &bla[0];
This is guaranteed to work since C++11, and effectively worked on all popular implementations before C++11.
If you need just a const char*, you can use either std::string::c_str() or std::string::data()
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3I know C++11 requires the bytes to be contiguous. But I don't see any requirement of the string pointed to by &bla[0] to be zero terminated. That may or may not be important to the user. Commented Dec 25, 2012 at 0:20
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3@user515430: The requirement is there, it is just well hidden. See this question.– MankarseCommented Dec 25, 2012 at 0:44
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1@Mankarse Given the number of comments back and forth, none of them from committee members, I'm not 100% convinced. But why use &bla[0] instead of c_str()? Modifying the std::string using the pointer &bla[0] is just ugly. Commented Dec 25, 2012 at 1:01
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I used the method described in the answer and it has worked fine in a function however i'm using it in a different function (pretty much exactly the same) and the output I get is rather strange. The output I get is
¡║½½½½½½½½½½½½½½EL)¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■¯■B¶↑h4ë¡║– RyanCommented Dec 25, 2012 at 1:30 -
1Using the internal
char*buffer of the string is perfectly fine and the prime way of calling e.g. Win32 API functions without allocating arrays ofcharyourself. What you have to do, is make sure you allocate enough space in thestring. As for "proof", see this answer. If quotes from the standard plus logical reasoning cannot prove this for you, I give up. Also: stop usingstrtok, especially if you havestd::stringwith its numerousfindfunctions andstd::stringstream.– rubenvbCommented Dec 25, 2012 at 9:39
const char*wouldn't it? But I'm guessing that's probably good enough.