How to assign a pointer to a list and print it out in c?

Question

I've managed to point towards an array, however, the closest I've gotten to pointing to a char* list only prints out the individual letters as shown below:

#include <stdio.h>

main()
{
char* list[2];
list[0] = "Hullo";
list[1] = "GoodBye";
char* pointer = &(list);
char* back = *(list);
printf("%c", back[8]);
}

0 through 4 of back prints out "Hullo", 5 through 7 is whitespace, and 8 through 14 prints out "Goodbye".

I'm wondering how I can avoid printing this list character to character, as it becomes a very inconvenient issue when returning lists of unspecified sizes and planning on using them for another function etc.

Answer 1

It feels like you're looking for:

#include <stdio.h>

int
main(void)
{
        char * list[2];
        list[0] = "Hullo";
        list[1] = "GoodBye";
        char ** pointer = list;
        for( size_t i = 0; i < sizeof list / sizeof *list; i++ ){
                printf("%s\n", *pointer++);
        }
}

but maybe you're looking for something like:

#include <stdio.h>
#include <stdlib.h>

char **
foo(void)
{
        char **p = malloc(3 * sizeof *p);
        if( p == NULL ){
                perror("malloc");
                exit(1);
        }
        p[0] = "Hullo";
        p[1] = "GoodBye";
        p[2] = NULL;
        return p;
}

int
main(void)
{
        char **p = foo();
        for( ; *p; p++ ){
                printf("%s\n", *p);
        }
        free(p);
}