Question

int
main(int argc,char **argv){

for (argc--, argv++; argc > 0; argc -= argCount, argv += argCount) {
      argCount = 1;
      switch (argv[0][1]) {
      case 'q':
        testnum = atoi(argv[1]);
        argCount++;
        break;
      default:
        testnum = 1;
        break;
      }
    }
//...............

my question is what does the argv[0][1] mean and the condition in for() confused me i mean for (argc--, argv++; argc > 0; argc -= argCount, argv += argCount)

//thanks guys….**argv[0][1] should be argv[0][1],thats my mistake not the code writers.

Answer 1

argv[0] represents the name of the program as it was invoked on the command line. If you typed ./myprogram --help, then argv[0] would be “./myprogram”.

argv[0][1] will be the second character of that string, ‘/’ in the example above.

---

Let’s see that for (argc--, argv++; argc > 0; argc -= argCount, argv += argCount):

It initializes the loop by doing argc-- then argv++ (argv now points to the second user parameter string) and argc declares an argument less.

The loop is for all arguments argc>0, and at every iteration, the number of treated arguments argCount is taken off the number of all arguments argc. That makes sense.

However switch (**argv[0][1]) doesn’t make any sense, argv[0][1] is a char, as seen before, not a pointer, so it cannot be dereferenced.