What does char * argv[] means?

The parameter char * argv[] decays to a pointer, char ** argv. You can equally well write the function signature for main() as:

int main(int argc, char ** argv)

You can do what you like with the pointer argv within main(), so argv++ for example just bumps argv to point at argv[1] rather than argv[0].

argv ---> argv[0] ---> "program"          argv[1] ---> "arg1"          argv[2] ---> "arg2"           ...          ...          argv[argc] == NULL

argv is an array of char*. Doing ++argv means accessing the next cell of the array. The * indicates we want the value of the cell, not the address.

When a program starts, it gets it's argument in the main function. That's why you ususally write.

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

This simply means that argv is a pointer to as many argument strings as indiciated by argc (== argument count). Since argv decays to char **argv you can also increase it, or you it otherwise like a pointer.

So if you want to print all arguments from the commandline:

int main(int argc, char *argv[]){   for(int i = 0; i < argc; i++)       printf("%s\n", argv[i]);   for(int i = 0; i < argc; i++)       printf("%s\n", argv++);    return 0;}