Possible Duplicate:
Convert std::string to const char* or char*
void setVersion(char* buf, std::string version) { buf = version; } I'm trying to write the version string into the buf, but the code above gave me this error "cannot convert ‘std::string {aka std::basic_string}’ to ‘char*’ in assignment".
What is the simplest way to fix it?
Assuming buf is at least version.length() + 1 bytes in size:
strcpy(buf, version.c_str()); 
First, there's a serious problem with the interface, since you don't know how large buf is. Without knowing this, there is no way you can correctly write anything to it. If you're passed the length, you can do something like:
void setVersion( char* buffer, size_t size, std::string const& version ) { size_t n = version.copy( buffer, size - 1 ); // leave room for final '\0' buffer[ n ] = '\0'; } Another possibility is that the intent is for you to set some global pointer; the given interface can't do this, since you have a copy of the pointer, but if you were given a reference to it, you might do:
void setVersion( char*& buffer, std::string const& version ) { if ( buffer != NULL ) { delete [] buffer; } buffer = new char[ version.size() + 1 ]; size_t n = version.copy( buffer, std::string::npos ); buffer[ n ] = '\0'; } (Other versions are possible, but you have to copy the string, to avoid lifetime issues.)
make sure buf has enough space to store the string
strcpy(buf,version.c_str()) Try adding a call to c_str() after version:
void setVersion(char* buf, std::string version) { buf = version.c_str(); } 
const char *.If all you want is a char* of the std::string - as pointed out earlier use vesrion.c_str().
If you do want to copy it into a separate buffer - for the function signature you have used - buffer should be allocated before the function is called and should be of same size as the version.size() + 1.
Otherwise you could do the following:
void setVersion ( char** out, std::string in ) { *out = new char[in.size() + 1]; strcpy ( out, in.c_str() ); } HTH
