semi-noob question

yawningdog · 09-09-2009, 01:20 PM

In college as a networking concentration, I took a few programming classes, mostly c++. Now I'm experimenting with a program in c that declares main like this.

Code:

int main(int argc, char *argv[])

In my programming classes and self study, I have never passed arguments to main. Can someone please explain the nature of these arguments? I suspect they have to do with command line arguments used when starting the program. If you need more info from me or if the question needs clarification, please let me know.

Thanks in advance.

johnsfine · 09-09-2009, 01:28 PM

Quote:

Originally Posted by yawningdog

they have to do with command line arguments used when starting the program.

Correct.

argv[0] is the command itself: the program name after whatever modification (path etc.) was done by the shell.

argv[1] is the first command line argument, etc.

argc is the count of them.

The best way to understand more is to write a simple C program that displays argv[0] through argv[argc-1] each on its own line of output. Then run that program with various input.

mobinskariya · 09-09-2009, 01:31 PM

i think argc represents argument count and *argv[] represents the parameters passed..that means argv[1] is the first argument, argv[2] the second one and so on..
edit:as johnsfine stated argv[0] represents the command itself.

yawningdog · 09-09-2009, 01:52 PM

Excellent. Thanks guys.