Array of pointers whose elements point to another array of pointers - arrays

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.

Answer 1

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]);
}
}