how to operate on global variables in c++ - c++

Question

I need help with global vars in C++. I am passing an empty argv array into a parse function and I am trying to fill it, however when I access argv back in main, argv[0] comes back empty no matter what I enter in. Please help.
static int argc;
static char* argv[100];
void parse(string str, int argc, char* argv[])
{
argc = 0;
char * cstr = new char [str.length()+1];
strcpy (cstr, str.c_str());
char * p = strtok (cstr," ");
int i = 0;
while (p!=0)
{
argc++;
argv[i] = p;
p = strtok(NULL," ");
i++;
}
delete[] cstr;
}
int main( void )
{
string theLine;
while (true)
{
cout << "myshell> ";
getline(cin,theLine);
cout << "Command was: " << theLine << endl;
if ((theLine == "exit") || (cin.eof()) || (theLine == "quit"))
exit(0);
parse(theLine, argc, argv);
cout << "argv[0] " << argv[0] << "\n";
}
}
I am trying to assign character strings to argv inside the parser function but when I go to access it in MAIN, it comes out blank. How do I achieve what I am trying to do? It has to do with global variables but I cannot figure out how.

Answer 1

The problem is that you're storing pointers into cstr in argv, but then you're deleting cstr at the end of the parse() function. Get rid of:
delete[] cstr;
Also, you should pass argc by reference. Otherwise, when you update it in the parse() function it won't update the caller's variable. So it should be:
void parse(string str, int &argc, char *argv[])
However, you also have a memory leak. You're creating a new cstr each time parse is called, but never deleting it. You should change the function to return cstr, and have the caller delete it when they're done with that parsed line.
char *parse(str str, int &argv, char *argv[]) {
...
return cstr;
}
Then in your main() loop, do:
int main( void )
{
string theLine;
while (true)
{
cout << "myshell> ";
getline(cin,theLine);
cout << "Command was: " << theLine << endl;
if ((theLine == "exit") || (cin.eof()) || (theLine == "quit"))
exit(0);
char *cstr = parse(theLine, argc, argv);
cout << "argv[0] " << argv[0] << "\n";
delete[] cstr;
}
}

Answer 2

You are confused about how global and local variables of the same name interact.
Let me start by saying that there are two distinct objects each named argc and two distinct objects each named argv.
The first set are the global variables
static int argc;
static char* argv[100];
The second set are the parameters of your parse function
void parse(string str, int argc, char* argv[])
{
argc = 0;
...
}
I've taken one line from parse. What that does is actually set the object in the argc parameter to 0. The object in the global argc is not set to 0 by this.
Let's rename some of the variables in the above snippet to see clearly which name belongs to which object/place:
static int global_argc;
static char* global_argv[100];
void parse(string str, int argc, char* argv[])
{
argc = 0;
...
}
Now, notice how I changed the first two to global_, but the rest (all of the, even the ones not shown) remain argc and argv.
You need to start by learning about variable scope. Once arrays and pointers are involved, such as argv, things get even more complicated. I could not possibly explain all of this to you in one answer, except for: first learn more about variables and variable scope. Then learn about pointers and arrays. Then come back to this program and see if you can fix it, which is not something you can do at this point in time.
It may take a little more time, and a little more work, but don't get discouraged.