Array of pointers whose elements point to another array of pointers

Question

What I need very precisely is an array A[10] and each of its element pointing to the respective element of array B[10] whose each element store its index.

Hence, A[1] points to B[1] and B[1] has value of 1. So, when I call *A[1] or *B[1], I get 1.

I know it can be super easy if the array B[10] is not an array of pointers but of integers but I need this for another purpose.

This is what I did but segmentation fault was offered.

#include <stdio.h>

int main() {
    int *A[10];
    int *B[10];
    
    for(int i=0; i<10; i++) {
        A[i] = B[i];
        *B[i] = i;
        printf("\n%d %d",*A[i],*B[i]);
    }
}

By the way, I am not very proficient in pointers.

Cannot reproduce, please provide a minimal reproducible example — Jabberwocky
– Jabberwocky, Commented Jun 9, 2021 at 7:03
@mr.loop your code works fine. Provide complete code that can be compiled. — Jabberwocky
– Jabberwocky, Commented Jun 9, 2021 at 7:06
x = *y[1]; is undefined behavior. You are de-referencing an uninitialized pointer. — Lundin
– Lundin, Commented Jun 9, 2021 at 7:27
@Jabberwocky My bad that I messed up the question. Edited finally to make it precise and reproducible. — mr.loop
– mr.loop, Commented Jun 9, 2021 at 17:14

Jabberwocky · Accepted Answer · 2021-06-10 06:03:16Z

3

Your commented code :

int main() {
    int *A[10];   // an array of 10 pointers, each of them pointing nowhere
    int *B[10];   // an array of 10 pointers, each of them pointing nowhere

    // now each array a and b contain 10 uninitialized pointers,
    // they contain ideterminate values and they point nowhere
    
    for(int i=0; i<10; i++) {
        A[i] = B[i];     // copy an uninitialized pointer
                         // this usually works but it's pointless

        *B[i] = i;       // you assign i to the int pointed by *B[i]
                         // but as *B[i] points nowhere you end up with a segfault

        printf("\n%d %d",*A[i],*B[i]);  // you never get here because the previous
                                        // line terminates the program with a segfault,
                                        // but you'd get a segfault here too for 
                                        // the same reason
    }
}

Your program is basically equivalent to this:

int main() {
    int *a;     // a is not initialized, it points nowhere
    *a = 1;     // probably you'll get a segfault here
}

Accessing the thing pointed by a pointer is called dereferencing the pointer. Dereferencing an uninitialized pointer results in undefined behaviour (google that term), most likely you'll get a seg fault.

I'm not sure what you're trying to achieve, but you probably want something like this:

#include <stdio.h>

int main() {
  int* A[10];
  int B[10];

  for (int i = 0; i < 10; i++) {
    A[i] = &B[i];
    B[i] = i;
    printf("%d %d\n", *A[i], B[i]);
  }
}

edited Jun 10, 2021 at 6:03

answered Jun 9, 2021 at 21:56

Jabberwocky

51.3k18 gold badges71 silver badges127 bronze badges

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

2 Comments

mr.loop Over a year ago

Your answer is correct. But is there a way to do this using pointer array rather than int array. Its because some of the elements of B have to store index and other point somewhere.

Jabberwocky Over a year ago

@mr.loop a way to do what? This is an XY Problem. You should tell us what you are actually trying to do by asking an other question. The problem with your code is that the pointers in the second array point nowhere.

Collectives™ on Stack Overflow

Array of pointers whose elements point to another array of pointers

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related