Why is my randomly generated array outputting only zeroes? - c++

Question

I am writing a code that is supposed to act as a lottery. The lottery numbers available are 1-50, and there are 10 of them. The user is supposed to input one number and the program returns if the user's number matches one of the 10 random lottery numbers. I have gotten all of these parts down, but there is one problem; all 10 of the lottery numbers must be unique. I have gotten 10 unique numbers 1-50, but they weren't very random. The code I have written up to this point seems correct to me, except I know there is something missing (along with the fact that I can clean my code up a lot, but I'm focusing on the objective right now). Right now if I run the program it will return ten zeroes. I need each element in the lottery array to be a unique number from 1-50, and produce different sets of numbers each time i run the program. Any help would be appreciated.
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <algorithm>
using std::cout; using std::cin; using std::endl;
void printOut(int[]);
void draw(int[]);
bool check();
int fillFunc[10];
int main()
{
const int arraySize = 10;
int win[arraySize] = {};
srand((unsigned)time(NULL));
draw(win);
cout << "Your lottery numbers are: ";
printOut(win);
}
void draw(int fillFunc[])
{
int i;
for (i = 0; i < 10; i++)
{
if (check() == true)
continue;
fillFunc[i] = 1 + rand() % 50;
}
}
void printOut(int fillFunc[])
{
for (int i = 0; i < 10; i++)
{
cout << " " << fillFunc[i];
}
cout << "\n";
}
bool check()
{
for (int i = 0; i < 10; ++i)
{
if (fillFunc[i] == i)
return true;
}
return false;
}
(also don't ask me why the array has the name "win", this is what my prof wants me to call it)

Answer 1

Because check() always return true. At the start, the array is filled with zeroes so check returns true as win[0]==0 and that remains true forever as nothing changes that value. So in draw you always branch to continue and never modify anything.
To solve the problem one way could be to shuffle the sequence of 1-50 and extract the first 10 values of the shuffled array. You can use the very simple Fisher-Yates algorithm.

Answer 2

To answer your question that you updated in the comments you need to consider a shuffling algorithm.
I will give you an idea of how to do this that is O(n) so you don't have to "loop" thru your current list of numbers and keep checking to see if the new number was already picked...
your lottery max number is 50 so make an array of size 50 that is as follows:
lotteryNumber[0]=1
lotteryNumber[1]=2
...
lotteryNumber[49]=50
to pick a number to put into your "selected" numbers array...
indexToLotteryNumbers = rand() % 50 - numbersPickedSoFar;
randomLotteryNumber[i++] = lotteryNumber[ indexToLotteryNumbers ];
// this is the key "trick"
swap(&lotteryNumber[ indexToLotteryNumbers ], &lotteryNumber[49-numbersPickedSoFar]);
numbersPickedSoFar++;
Theory
The random numbers are the indexes and not the actual values
By swapping the number you picked with the outer element you don't care if your random number generator picks the same index because it will be different the next time.
look at a small example...say you have 1 2 3 4
rand() generates "2"
so your first # is 2 (assume 1-based indexing here) now you swap the (4th) element for the (2nd) element now you have 1 4 3 |2
(here |2 means you can't pick that number again because it's outside the
random # generator range) but you reduce the random # generation from
1-4 to 1 thru 3
rand() generates "1" so your number is "1" and you swap that with the "outer" unpicked number 3 4 |1 2
rand() generates
"2" again! your lottery number is 4 this time... and so on.
I hope this makes sense.

Answer 3

This is what I would do, given your constraints. Rather than check if a number is unique while you are filling the array, just pass the array into the function that chooses the number so it can return a unique value.
I also removed the redundant global array. It could be a source of bugs if you forgot to pass the local array you were working with to any of the functions.
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
int getUniqueNumber(int fillFunc[])
{
while(true)
{
//pick number
int val = 1 + rand() % 50;
//assume it's unique
bool unique = true;
for (int i = 0; i < 10; ++i)
{
//if another number matches, it isn't unique, choose again
if (fillFunc[i] == val)
{
unique = false;
break;
}
}
//if it is unique, return it.
if (unique)
{
return val;
}
}
//never reached, but avoids an all control paths must return a value warning.
return -1;
}
void draw(int fillFunc[])
{
for (int i = 0; i < 10; i++)
{
fillFunc[i] = getUniqueNumber(fillFunc);
}
}
void printOut(int fillFunc[])
{
for (int i = 0; i < 10; i++)
{
cout << " " << fillFunc[i];
}
cout << "\n";
}
int main()
{
srand((unsigned)time(NULL));
const int arraySize = 10;
int win[arraySize] = {};
draw(win);
cout << "Your lottery numbers are: ";
printOut(win);
return 0;
}
There are other, perhaps better, ways to select unique numbers in a range, but I went for simple to implement and explain. You can read about some other methods in these questions:
Unique random numbers in an integer array in the C programming language
Unique (non-repeating) random numbers in O(1)?

Answer 4

You only assign a value to fillFunc[i] when i is 10, which is an out of bounds access. Move the assignment to inside the loop.
You have other problems too. If you do continue, you just leave an entry in the array unset.
You should try adding a lot of output statements to your program so you can more easily understand what it's doing. If you prefer, use a debugger to step through it.