How is the ARGC value initialized if I don't enter it in the command prompt?
<p>When I run the following code: </p>

<pre><code>void main(int argc, char** argv)
{

    if (argc != 2){
        fprintf(stderr, "Usage: %s video-dir-path\n", argv[0]);

        exit(-1);
    }


    XFishTracker ft(argv[1]);

    int id = 0;

    while (true)
    {
        id++;
...
...
</code></pre>

<p>It exits since the value of ARGC is obviously NOT 2. 
If I comment out the exit (-1) line, I get an ASSERTION ERROR. 
I think because the ARGC isn't 2, my program does not run or proceed. How do I initialize argc to 2 and make the program run, when it exits before I can even see the command prompt properly. </p>

<p>How do I make the command prompt stay and give two inputs so that argc == 2?</p>