Getting compiler error when trying to print out contents of vector

Question

Here's what I have so far:

 int main(int argc, char const *argv[])
    {
        srand(time(NULL));
        vector<int> deck;
        vector<bool> drawn;
        for (int i = 0; i < 20; i++)
        {
            int numvalue = rand()%20;

            if (drawn[numvalue - 1])  // this checks if their are duplicate values
            {
                i--;
                continue;
            }
            else
            {
                drawn[numvalue - 1] = true;
                //vector<int> deck;
                deck.push_back(numvalue);
            }
        }
        copy (deck.begin(), deck.end(), ostream_iterator<int>(cout, " "));
        return 0;
    }

the Compiler error I'm getting is:

Edit: solved the compiling issue, but now this is seg faulting...

I'd like to print out the contents inside my vector deck, and after doing some reading I thought the best option would be to use the function copy. Any ideas how I can fix this?

Answer 1

Declare deck at the top of main like such, so it's not declared out of scope

int main(int argc, char const *argv[])
{
    srand(time(NULL));
    vector<int> deck;
    vector<bool> drawn(20);
    for (int i = 0; i < 20; i++)
    {
        int numvalue = rand()%20;

        if (drawn[numvalue - 1])
        {
            i--;
            continue;
        }
        else
        {
        drawn[numvalue - 1] = true;
        deck.push_back(numvalue);
        }
    }
    copy (deck.begin(), deck.end(), ostream_iterator<int>(cout, " "));
    return 0;
}

Also

int numvalue = rand()%20;

can easily return 0, and with if (drawn[numvalue - 1]), you're accessing drawn[-1] about 5% of the time. Much smarter to change it to:

int numvalue = rand()%20 +1;

Also as WhozCraig points out below, you never initialized drawn, which will lead to segfaults.

Answer 2

Your deck is being defined in a scope that has finished by the time the copy is called. Move the declaration of deck up further before the start of the for loop so it remains in scope.