# While loops and arrays

Printable View

• 05-14-2010
Freshjunior
While loops and arrays
Hello, I was going through a beginner's guide book on C++ and could not understand exactly how this example works, even after pondering for hours.

This program is a card-dealing program which uses a 52-element array in which each element corresponds to a card. Once a card is picked, this array is updated to show that the card should be skipped in the future.

Code:

```#include <iostream> #include <stdlib.h> #include <time.h> #include <math.h> using namespace std; int rand_0toN1(int n); void draw_a_card(); int select_next_available(int n); char *suits[4] =  {"hearts", "diamonds", "spades", "clubs"}; char *ranks[13] = {"ace", "two", "three", "four", "five",                 "six", "seven", "eight", "nine",                 "ten", "jack", "queen", "king" }; int card_drawn[52]; int cards_remaining = 52; int main() {     int n, i;     srand(time(NULL));      // Set seed for random numbers.     while (1) {         cout << "Enter no. of cards to draw (0 to exit): ";         cin >> n;         if (n == 0)             break;         for (i = 1; i <= n; i++)             draw_a_card();     }     return 0; } // Draw-a-card function // Performs one card-draw by getting a random 0-4 and a random //  0-12. These are then used to index the string arrays, ranks //  and suits. // void draw_a_card() {     int r;        // Random index (0 thru 12) into ranks array     int s;        // Random index (0 thru 3) into suits array     int n, card;     n = rand_0toN1(cards_remaining--);     card = select_next_available(n);     r = card % 13;            // r = random 0 to 12     s = card / 13;            // s = random 0 to 3     cout << ranks[r] << " of " << suits[s] << endl; } // Select-next-available-card function. // Find the Nth element of card_drawn, skipping over all those //  elements already set to true. // int select_next_available(int n) {     int i = 0;     // At beginning of deck, skip past cards already drawn.     while (card_drawn[i])         i++;     while (n-- > 0) {    // Do the following n times:         i++;                      // Advance to next card         while (card_drawn[i])      // Skip past cards             i++;                  //  already drawn.     }     card_drawn[i] = true;        // Note card to be drawn     return i;                    // Return this number. } // Random 0-to-N1 Function. // Generate a random integer from 0 to N-1. // int rand_0toN1(int n) {     return rand() % n; }```
I mainly have problems with the select_next_available card function.
1.for the " while (card_drawn[i]) ", how is the condition card_drawn[i] used? The initial value of i is 0, so card_drawn[0] is false and the loop shouldnt even start in the first place?

2.For the while (n-- > 0) loop, does that mean n is further reduced by one here? What is the purpose of doing n--, and if n is initially zero (the rand_0toN1 function picks 0 from the numbers 0 to 51), then n-- = -1??

3.Lastly, why is i increased so many times in this function? I can see i++ three times here.

This example is really getting on my nerves~

Thanks!
• 05-15-2010
jdragyn
1. The value of card_drawn[i] is not 0 simply because i is 0. Whatever is in the i-th element of card_drawn[] is what the while() is testing.

2. because the -- is after the n, the value of n is used first, and THEN it is decremented by 1. Since n is passed in to the function, n could be anything. This loop effectively counts from that n down to 0. (or as we'll see below, skips over n undrawn cards)

3. i is increased several times for different purposes. The first while loop does what the comment states, so if it skips past 5 cards, i == 5 now (remember i started at 0). Next it advances through n undrawn cards (it skips over any card in that while(n--...) loop that was already drawn by doing the same thing that the first while loop did). When it is done, i has been incremented several times to the target value, which is then returned.
• 05-15-2010
Freshjunior
I understand that the part while (n-- >0) means do the following n times.
But why does
Code:

```while (card_drawn[i])         i++;```
means finding the first available card by skipping past all the drawn cards? Is it because all the drawn cards are tagged "true"? so for example element 3 is true, while(card_drawn[2]) is true and so i = i+1 =3, the computer skips to element 4?
• 05-15-2010
Freshjunior
btw, how does return i work? What do you mean when you "return" this number. Does it mean returning i to the beginning of the function, so i becomes 0 once again?
• 05-15-2010
User Name:
return, for most calling conventions, places the return code(0, in this case) in eax and executes a ret. lol
Basically, it's telling the compiler to finish the function(or the whole program, as is the case if it's used in "int main()") and report the 0 for an error code, 0, in most cases, means everything went good. The WinAPI alternative is ExitProcess.

Actually, does anyone know if there's some black magic in the return 0 in int main. I know for a fact that compilers add extra code to parse argc and argv from Window's style of command line to any non-Linux executable(Linux actually pushes argc and argv onto the stack). I wonder if compilers do OS specific things after return 0 too.