Converting string to char** - c++

Question

I have a problem with my code and i dont know whats wrong :(
I want to convert a string to a char* array and print it out at the end.
The output is currently 3 times the last word of the sentence i entered.
void parse(std::string &s, char **argv)
{
std::istringstream iss(s);
std::string tmp;
while(iss >> tmp)
{
*argv++ = (char*) tmp.c_str();
}
}
int main()
{
std::string input;
while (1)
{
std::getline(std::cin, input);
int argCount = countArgs(input);
char *argv[argCount];
parse(input, argv);
for(int i=0; i<argCount; ++i)
{
std::cout << argv[i] << std::endl;
}
}
return 0;
}

Answer 1

I won't get into details why your current code does not work, because what you are doing is basically not-really-safe/good/sane thing. Please rethink your approach.
Why use char* if you can use string?
If possible, instead of
char *argv[argCount];
...
void parse(std::string &s, char **argv)
...
*argv++ = (char*) tmp.c_str();
use
string argv[argCount];
...
void parse(std::string &s, string* argv)
...
*argv++ = tmp;
it will work almost identically to what you have now, "just" using a different datatype.
However, note that this all will still not be really C++'y, as you are using string argv[argCount]; with non-constant argCount. I'm suprised your compiler actually compiles it, you must be using some C-compliance compilation flag. You should be using std::vector or std::list to keep an "array of variable length".

Answer 2

You can avoid the issues mentioned in the comments by calling strdup() before saving your char *:
void parse(std::string &s, char **argv)
{
std::istringstream iss(s);
std::string tmp;
while(iss >> tmp)
{
*argv++ = strdup( tmp.c_str() );
}
}
You'll need to free (not delete()) these copies later.
for (int i = 0; i < argCount; ++i)
{
std::cout << argv[i] << std::endl;
free( argv[i] );
}