C-向函数内的双指针分配值

Question

I'm trying to understand C, specifically double-pointers and i came across this problem. I know for a single pointer (with removing the for loop, etc) that this concept would work, but i seem to be getting a seg fault at the located comment.

Could someone explain why as to why i'm getting this error? I have a hunch that before i pass the address of myArgs i need to allocate some memory for it, but since i'm just doing a shallow copy, do i still need to allocate memory?

void readArgs(int argc, char *argv[], char ***myArgs) {
   int i;
   for(i = 0; i < argc; i++) {
      /* crashes here @ i = 0 */
      *myArgs[i] = argv[i];
   }
}

int main(int argc, char *argv[]) {
   char **myArgs;
   int i;

   readArgs(argc, argv, &myArgs);
   for(i = 0; i < argc; i++)
      printf("arg[%d]: %s\n", i, myArgs[i]);
}

Answer 1

You're getting a segfault because myArgs is uninitialized. You should indeed allocate space with malloc . You're not making a shallow copy; you're copying an array of pointers.

That said, triple pointers are a code smell in C. You should never need more than ** .