Optimal way to perform a shift operation on an array

Question

Suppose I have an array

unsigned char arr[]= {0,1,2,3,4,5,6,7,8,9};

Is there a way to perform shift operation on them besides just copying them

Answer 1

If you're looking for a pure C solution, here it is, including a driver program. It turns out to be pretty simple: to rotate by n, you:

1. reverse the first n elements in-place,
2. reverse the remaining elements in-place, and
3. reverse the whole array in-place.

This requires one element worth of extra storage (for reversing).

#include 
#include 
#include 

/* print an array */
static void print_array(unsigned char *arr, size_t n, const char *prefix)
{
    size_t i;

    if (prefix) {
        printf("%s: ", prefix);
    }
    for (i=0; i < n; ++i) {
        printf("%02x ", (unsigned int)arr[i]);
    }
    printf("\n");
}

/* reverse 'arr', which has 'narr' elements */
static void reverse(unsigned char *arr, size_t narr)
{
    size_t i;

    for (i=0; i < narr / 2; ++i) {
        unsigned char tmp = arr[i];
        arr[i] = arr[narr-i-1];
        arr[narr-i-1] = tmp;
    }
}

/* rotate 'arr' of size 'narr' by 'shift' */
static void rotate(unsigned char *arr, size_t narr, unsigned long shift)
{
    reverse(arr, shift);
    reverse(arr + shift, narr - shift);
    reverse(arr, narr);
}

/* driver program */
int main(int argc, char *argv[])
{
    unsigned char arr[]= {0,1,2,3,4,5,6,7,8,9,10};
    size_t narr = sizeof arr / sizeof arr[0];
    unsigned long shift = 2;

    if (argc > 1) {
        char *eptr;
        shift = strtoul(argv[1], &eptr, 0);
        if (*eptr || errno == ERANGE) {
            perror("strtoul");
            return EXIT_FAILURE;
        }
    }
    print_array(arr, narr, "before shift");
    rotate(arr, narr, shift);
    print_array(arr, narr, "after shift");
    return EXIT_SUCCESS;
}