c++ - Mixing two arrays by alternating elements two by two -

what elegant algorithm mix elements 2 two in 2 arrays (of potentially differing sizes) items drawn in alternating fashion each array, leftovers added end?

e.g.

array 1:        0, 2, 4, 6  array 2:        1, 3, 5, 7  mixed array:    0, 2, 1, 3, 4, 6, 5, 7 

don't worry null checking or other edge cases, i'll handle those. here solution not work properly:

for (i = 0; < n; i++) {      arr[2 * + 0] = a[i];      arr[2 * + 1] = a[i+1];        arr[2 * + 0] = b[i];      arr[2 * + 1] = b[i+1]; } 

it fiddly calculate array indices explicitly, if arrays can of different , possibly odd lengths. easier if keep 3 separate indices, 1 each array:

int pairwise(int c[], const int a[], size_t alen, const int b[], size_t blen) {     size_t = 0;       // index     size_t j = 0;       // index b     size_t k = 0;       // index c      while (i < alen || j < blen) {         if (i < alen) c[k++] = a[i++];         if (i < alen) c[k++] = a[i++];         if (j < blen) c[k++] = b[j++];         if (j < blen) c[k++] = b[j++];     }      return k; } 

the returned value k equal alen + blen, implicit dimension of result array c. because availability of next item checked each array operation, code works arrays of different lengths , when arrays have odd number of elements.

you can use code this:

#define countof(x) (sizeof(x) / sizeof(*x))  int main() {     int a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};     int b[] = {-1, -2, -3, -4, -5, -6};      int c[countof(a) + countof(b)];     int i, n;      n = pairwise(c, a, countof(a), b, countof(b));      (i = 0; < n; i++) {         if (i) printf(", ");         printf("%d", c[i]);     }     puts("");      return 0; } 

(the example in c, not c++, code doesn't use of c++'s containers such vector, i've uses plain old ´int` arrays explicit dimensions, same in c , c++.)