How to call another function that utilizes argc, char *argv[] in main without passing things to it?-CodePudding

say there is a function in my code: int getwords(int argc, char *argv[]) and I want to call this function in main(). How do I call this in main without erroring out?

`void main(void)
    getwords();`

CodePudding user response：

If you write main to provide access to argc and argv:

int main(int argv, char **argv) {
  // ...
}

Then you can pass those to another function when you call it:

void foo(int argc, char **argv) {
  // ...
}

int main(int argc, char **argv) {
  foo(argc, argv);
}

argc will be copied, but argv is a pointer, so the pointer is copied but not the data it points to.

CodePudding user response：

if for some special reason you do not want to use parameters you can use global variables available in all program.

int gargc;
char **gargv;

int main(int argc, char *argv[])
{
    gargc = argc;
    gargv = argv;

    getopt();
}