confused about this argv use - c++

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

That code doesn't look correct. **argv[0][1] tries to dereference a char.
argv[0][1] would make sense, and means "take the second char of the first char* in argv." IMHO, the code is trying to detect a -q command-line flag (and subsequently setting testnum to the int version of the next argument, blindly assuming it is present), but it's skipping checking for the -, and blindly assuming it's there, and no other arguments would ever have q as a second character.
This code needs to be refactored. Here's one way:
int main(int argc, char **argv) {
int testnum = 1;
for (int argi = 1; argi < argc; ++argi) {
if(argv[argi][0] == '-') {
switch (argv[argi][1]) {
case 'q':
if(argi + 1 == argc || argv[argi + 1][0] == '-') {
/* Handle missing argument error. */
return 1;
}
testnum = atoi(argv[++argi]);
break;
default:
/* Handle unrecognized flag error. */
return 1;
}
}
else
{
/* Handle non-flag parameter. */
}
/* Continue with program. */
return 0;
}

Answer 2

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.

Answer 3

This code looks very crazy. I guess you intended to do the following:
int main(int argc,char **argv){
char** p = argv + 1; // skipping program name
while (*p != 0) { // over all parameters
testnum = 1;
if (*p[1] == 'q') { // skipping - of "-q", not a good idea
p ++;
if (*p != 0) { // may be null, read testnum from parameter,
// ?? no check whether this is an integer at all
testnum = atoi(*p);
}
}
}
(not tested, may not compile nor work)