Command-line Parameters in C program?

How do I read command-line parameters in C? For example, in

./test --help


./test --build

how do I access "--build" or "--help"?

-------------Problems Reply------------

Your parameters are in argv:

int main(int argc, char **argv)

if you printf the content of argv (argv[0],argv[1] etc) youll get the idea.


int main (int argc, char **argv)
for(int i = 0;i< argc;i++)

You can use the argc and argv arguments to the main function and do different things based on them:

#include <string.h>
void dohelp(void) { /* whatever; */ }
void dobuild(void) { /* whatever; */ }
int main(int argc, char **argv) {
if (argc == 2) {
if (!strcmp(argv[1], "--help")) dohelp();
if (!strcmp(argv[1], "--build")) dobuild();
return 0;

argc contains the number of parameters passed by the shell to your program, including the program name. So myapp --help gets an argc of 2.

argv are the arguments themselves. The last argv (argv[argc]) is the NULL pointer.

Edit: the parameters don't need to be named argc and argv, but naming else something else is very, very bad!

int main(int foo, char **bar) /* RGAGGGGHH */
int main(int n, char **options) /* RGAGGGGHH */

The very basic is to use the arguments (int argc, char *argv[]) and you can parse those directly.

One more advanced method is to use getopt...

There are several ways to do it [as usual]. Command line arguments are read from argv (passed to main along with argc).

You can parse those yourself and have a bit switch setting flags each time a new option is found in argv. Or you can use a library to parser command line arguments. I suggest libc getopt (google it).

Category:c# Views:0 Time:2011-03-12
Tags: c# parameters gcc

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