Iterate through an array of pointer of pointers

Question

I want to iterate through **list_arg[].

I can't make this work, unless I try to iterate through it in another function, then I can do list_arg++.

int main(int argc, char *argv[]) {
    char *three[] = {"Ver", NULL}
    char *two[] = {"john", NULL};
    char *one[] = {"Majka", NULL};
    char **list_arg[] = { one, two, three, NULL};

   while (**list_arg != NULL) {
       printf("%s", **list_arg);
       list_arg++; //invalid increase
       //(*list_arg)++; iterates through one[] only.
       //(**list_arg)++; iterates through *one[], aka "Majka" only.
   }
   return 0;
}

@SupportUkraine so its not possible to iterate through it? What options/modifications could I do. — Olivia22
– Olivia22, Commented Jul 15, 2022 at 4:44

Adrian Mole · Accepted Answer · 2022-07-15 04:48:39Z

1

When using arrays of pointers (and especially arrays of pointers to pointers), it is generally better to use the specific [] operator, rather than trying to work out which combination of dereferencing (*) operators will get to (and manipulate) the required target.

The following code does what (I think) you want (I have added an extra name to one of the arrays, to show that the iteration does work):

#include <stdio.h>

int main()
{
    char* three[] = { "Ver", NULL };
    char* two[] = { "John", "Fred", NULL}; // Add extra one here for demo
    char* one[] = { "Majka", NULL };
    char** list_arg[] = {one, two, three, NULL};

    for (size_t i = 0; list_arg[i] != NULL; ++i) {
        char** inner = list_arg[i];
        for (size_t j = 0; inner[j] != NULL; ++j) {
            printf("%s ", inner[j]);
        }
        printf("\n"); // New line to separate inner arrays
    }
    return 0;
}

answered Jul 15, 2022 at 4:48

Adrian Mole

52.1k193 gold badges61 silver badges101 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Olivia22 Over a year ago

this works. thanks. Ill certainly remember that next time.

0___________ Over a year ago

@Olivia22 simple use pointer instead of an array and you will be able to use while loops to iterate through all level of indirection. See my answer

0___________ · Accepted Answer · 2022-07-15 08:12:29Z

You can iterate if you use pointer instead array.

int main(int argc, char *argv[])
{
    char *three[] = {"Ver", NULL};
    char *two[] = {"john", NULL};
    char *one[] = {"Majka", NULL};
    char ***list_arg = (char **[]){ one, two, three, NULL};

    while(*list_arg)
    {
        while(**list_arg)
        {
            while(***list_arg)
            {
                fputc(***list_arg, stdout);
                (**list_arg)++;
            }
            putc('\n', stdout);
            (*list_arg)++;
        }
        list_arg++;
    }

   return 0;
}

https://godbolt.org/z/Y3benvWja

or without compound literals:

int main(int argc, char *argv[])
{
    char *three[] = {"Ver", NULL};
    char *two[] = {"john", NULL};
    char *one[] = {"Majka", NULL};
    char **list_arg_array[] = { one, two, three, NULL};
    char ***list_arg = list_arg_array;

    while(*list_arg)
    {
        while(**list_arg)
        {
            while(***list_arg)
            {
                fputc(***list_arg, stdout);
                (**list_arg)++;
            }
            putc('\n', stdout);
            (*list_arg)++;
        }
        list_arg++;
    }

   return 0;
}

Collectives™ on Stack Overflow

Iterate through an array of pointer of pointers

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related