Question

This is a learning exercise. I’m attempting to augment memcpy by notifying the user if the copy operation will pass or fail before it begins. My biggest question is the following. If I allocate two char arrays of 100 bytes each, and have two pointers that reference each array, how do I know which direction I am copying? If I copy everything from the first array to the second how do I ensure that the user will not be overwriting the original array?

My current solution compares the distance of the pointers from the size of the destination array. If the size between is smaller than I say an overwrite will occur. But what if its copying in the other direction? I’m just kind of confused.

int memcpy2(void *target, void *source, size_t nbytes) {
    char * ptr1 = (char *)target;
    char * ptr2 = (char *)source;


    int i, val;

    val = abs(ptr1 - ptr2);
    printf("%d, %d\n", val, nbytes + 0);
    if (val > nbytes) {
        for (i = 0; i < val; i++){
            ptr1[i] = ptr2[i];
        }
        return 0;  /*success */
    }
    return -1; /* error */
}



int main(int argc, char **argv){

  char src [100] = "Copy this string to dst1";
  char dst [20];
  int p;
  p = memcpy2(dst, src, sizeof(dst));

    if (p == 0)
        printf("The element\n'%s'\nwas copied to \n'%s'\nSuccesfully\n", src, dst);
    else
        printf("There was an error!!\n\nWhile attempting to copy the elements:\n '%s'\nto\n'%s', \n Memory was overlapping", src, dst);
    return 0;


}

Answer 1

The only portable way to determine if two memory ranges overlap is:

int overlap_p(void *a, void *b, size_t n)
{
    char *x = a, *y =  b;
    for (i=0; i<n; i++) if (x+i==y || y+i==x) return 1;
    return 0;
}

This is because comparison of pointers with the relational operators is undefined unless they point into the same array. In reality, the comparison does work on most real-world implementations, so you could do something like:

int overlap_p(void *a, void *b, size_t n)
{
    char *x = a, *y =  b;
    return (x<=y && x+n>y) || (y<=x && y+n>x);
}

I hope I got that logic right; you should check it. You can simplify it even more if you want to assume you can take differences of arbitrary pointers.