case 2
just prints the first 5 cards, without removing them from the deck...
So dealing a new time (without shuffle) will still return the same cards.
[OT] deckArray
may be rewritten as follow:
vector<node> deckArray()
{
vector<node> cards(52);
const char* colors[4] = {"S", "H", "D", "C"};
int i = 0;
for (int v = 2; v != 15; ++v) {
for (int c = 0; c != 4; ++c, ++i) {
cards[i].number = v;
cards[i].suit = colors[c];
}
}
random_shuffle(cards.begin(), cards.end());
return cards;
}