Sure, let me just (s0 ^= b[i + 0]), (s1 ^= b[i + 1]), (s2 ^= b[i + 2]), (s3 ^= b[i + 3]);
          ({ s0, s1, s2, s3 } = encrypt(xk, s0, s1, s2, s3));
          (o[i++] = s0), (o[i++] = s1), (o[i++] = s2), (o[i++] = s3); for you.