why is argc returning 6 as opposed to 3?

Question

I have following program and i am passing it 2 arguments at the command line as shown below. I was expecting the argc to be 3, but it prints it as 6. why?

#include <iostream>
#include <string>
#include <stdlib.h>

using namespace std;
void usage();

int main(int argc, char *argv[])
{
    cout << argc << endl;
    if (argc != 3)
            usage();

    string regex = argv[1];
    string searchString = argv[2];

    cout << "regex: " << regex << " . searchString: " << searchString << endl;

    return 0;
}

void usage()
{
    cout << "Usage: ./stringmatch <regex> <searchString>" << endl;
    exit(1);
}

Command line:

[jim@cola c++]$ ./stringmatch [yyyy\d\d\d]* yyyy1234
 6
 Usage: ./stringmatch <regex> <searchString>

Answer 1

Your shell is expanding the glob pattern [yyyy\d\d\d]* so the actual number of arguments this results in depends on the contents of the current directory!

The [yyyy\d\d\d] becomes a character class matching the characters y and d, and the * matches anything that follows, so I'm guessing your current directory has 4 files that start with y or d. To see what it expands to, use echo:

$ echo [yyyy\d\d\d]*

To make it work as intended, quote the argument:

$ ./stringmatch '[yyyy\d\d\d]*' yyyy1234