What am I doing wrong with srand to build a non repeating random array in C++?

Question

I am trying to build a random array of length (size1). The way that I've researched doing this is to have two separate arrays, one for my random numbers and a secondary "checking" array to make sure the numbers don't repeat. These are labeled (shuffle) and (visit) in my code respectively. count1 is an integer for counting through my for loop.

I have included the following in different combinations and it hasn't worked.

    #include <ctime>
    #include <time.h>
    #include <cstdlib>

The code I seem to be struggling with is this:

    srand((unsigned)time(0));

for (count1 = 0; count1 < size1; count1++)
{
    num = (rand()%size1);
        if (visit[num] == 0)
        {
            visit[num] = 1;
            shuffle[count1] = num;
        }
}

Answer 1

It is easier to fill your array with the numbers 0 to size1-1 and then shuffle those numbers.

So in c like code (haven't used c in a long time):

for (int count = 0; count < size1; count++) {
    shuffle[count] = count;
}

and then shuffle it

for (int index = 0; index < size1; index++) {
    int shuffleIndex = index + rand() % (size1 - index);
    int temp = shuffle[shuffleIndex];
    shuffle[shuffleIndex] = shuffle[index];
    shuffle[index] = temp;
}

Basically, for the first element in the array, select any index and switch the values. Now the first index is randomly selected. For the second element, select a random index (excluding the first as that has already been randomly selected) and do the same. Repeat until you get to the last element.

Answer 2

There's nothing wrong with your usage of srand. But, the task you describe is very simple, and doesn't involve the use of the rand function.

std::vector<int> v(size);
std::iota(v.begin(), v.end(), 0);
std::random_shuffle(v.begin(), v.end());

If you don't have the iota function, it just generates an incrementing sequence of integers.

Answer 3

Your #includes are not the problem. The problem is: you are not trying to generate alternate numbers when the first one you generate has already been used.

You should include a while loop, inside the for loop, which keeps generating new numbers until you find one that works.

They way it's written now, if you generate the sequence 4,1,4,2,4 for size1 = 5, your array will end up looking like this:

{4,1,0,2,0}

Assuming each entry was set to 0 initially. This is because you will just skip over the indices for which the two extra 4s were generated.