Shallow copy of two arrays in C - c

Question

Is there a way to shallow copy the elements of a dynamically allocated array in C? Something like the following:
int size = 2;
int *arr1 = malloc(size*sizeof(int));
int *arr2 = malloc(size*sizeof(int));
arr1[0] = 1; arr1[1] = 2; // {1, 2}
arr2[0] = 3; arr2[1] = 4; // {3, 4}
// shallow copy here s.t. memadress(arr1[i]) == memadress(arr2[i])
// ...
arr1[0] = -1; // arr1 = {-1, 2} AND arr2 = {-1, 4}

Answer 1

"Shallow" or "soft" copy typically just means copying a pointer but not the pointed-at data. As opposed to "hard" copy which also copies the pointed-at data. In your case, for example:
int *arr1 = malloc(size*sizeof(int));
int *arr2 = arr1; // "soft copy"
int* arr3 = malloc(size*sizeof(int);
memcpy(arr3, arr1, size*sizeof(int)); // "hard copy"
Or if you will:
int *arr1 = malloc(size*sizeof(int));
int *arr2 = malloc(size*sizeof(int));
arr1[0] = 1; arr1[1] = 2; // {1, 2}
arr2[0] = 3; arr2[1] = 4; // {3, 4}
free(arr2);
arr2 = arr1;
arr1[0] = -1; // {-1, 2} AND arr2 --> {-1, 4}
This doesn't make much sense, since it implies a soft copy of the elements not of the arrays. Such a container is likely needlessly complicated and not very useful. It could be done with an array of pointers, however:
#include <stdio.h>
#include <stdlib.h>
int main (void)
{
int size = 2;
int **arr1 = malloc(size*sizeof(int*));
int **arr2 = malloc(size*sizeof(int*));
// assign pointers to point at local compound literals:
arr1[0] = &(int){1}; arr1[1] = &(int){2}; // {1, 2}
arr2[0] = &(int){3}; arr2[1] = &(int){4}; // {3, 4}
arr2[0] = arr1[0]; // "soft copy"
*arr1[0] = -1;
printf("arr1: {%d %d}\n", *arr1[0], *arr1[1]);
printf("arr2: {%d %d}\n", *arr2[0], *arr2[1]);
free(arr1);
free(arr2);
}
Output:
arr1: {-1 2}
arr2: {-1 4}
But please avoid coming up with such obscure solutions just for the heck of it. Good programming = writing code as simple as possible.