Creating Random Number Using Vectors in C++

Question

I'm building a program in which a user types in a number (n) and a set of random numbers is created. So for example, if a user inputs 8, then eight random numbers should be created and they should range from 0-999,999. The program seems to be compiling, the only problem is, only one random number is being generated.

#include <iostream>
#include <vector>
#include <cstdlib> 

using namespace std;

main()
{
    int n;
    int r;
    int i;
    int j;
    vector<int> v;

    cout << "Enter size of vector: ";
    cin >> n;

    for (i = 0; i < n; i++)
    {
        v.push_back(n);
        r = rand() % 1000000;
        v[i] = r;
    }

    cout << r << endl;

Can anyone tell me what I'm doing wrong and what I need to do for more than one random number to be generated?

How do you know that only one number is being generated? Your cout statement at the end will only print one number. — GWW, Jan 29 '11 at 01:13

score 4 · Accepted Answer · edited May 23 '17 at 12:28

4

What's wrong with the obvious:

for (int i=0; i<n; i++)
    v.push_back(rand()%1000000);

It looks like you're generating the right quantity of random numbers, but when you're done, you're printing r instead of v, which is what contains the random numbers.

Edit: std::vector doesn't support operator<< directly, so you can use a loop to print out the contents:

for (int i=0; i<v.size(); i++)
    std::cout << v[i] << '\n';

or you can use std::copy:

std::copy(v.begin(), v.end(), std::ostream_iterator<int>(std::cout, "\n"));

There are, of course, a variety of other possibilities as well...

Edit 2: Here's a complete/correct version of what Chris Lutz suggested in his comment:

#include <vector>
#include <iostream>
#include <algorithm>
#include <iterator>
#include "infix_iterator.h"

template <typename T>
std::ostream& operator<<(std::ostream &o, const std::vector<T>& v) { 
    o << "[";
    std::copy(v.begin(), v.end(), infix_ostream_iterator<T>(o, ", ")); 
    o << "]";
    return o; 
}

#ifdef TEST
int main() { 

    std::vector<int> x;

    for (int i=0; i<20; i+=2)
        x.push_back(i);

    std::cout << x << "\n";
    return 0;
}
#endif

Though it's not strictly necessary, this uses an ostream_infix_iterator I posted some time ago.

edited May 23 '17 at 12:28

Community

1
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answered Jan 29 '11 at 01:14

Jerry Coffin

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@Jerry Coffin - I tried putting v in place of r in the cout statement but when I run the program, the only output is "0". – UndefinedReference Jan 29 '11 at 01:20
@Meursault- You shouldn't be able to print out a vector with `cout << v << endl;` and I'm surprised that's compiling at all. To print the vector, just use a for loop over the elements: `for (size_t i = 0; i < v.size(); ++v) cout << v[i] << endl;` – templatetypedef Jan 29 '11 at 01:22
@templatetypedef - You _can't_ print a vector with `cout << v << endl;` but you most certainly _should_ be able to do so. – Chris Lutz Jan 29 '11 at 01:28
@Chris Lutz- True... that was a poor choice of words. By "shouldn't be able to" I meant "this shouldn't compile on any major compiler," not "this isn't good functionality." I actually am curious why the spec doesn't have this functionality as part of the STL? – templatetypedef Jan 29 '11 at 01:29
@templatetypedef - You are right, I meant to write v[i], it will not compile with just v, and thanks, I changed the last bit of your for loop from ++V to i++ and it worked. Thank you! – UndefinedReference Jan 29 '11 at 01:31
@templatetypedef - I am as well. It seems mind-numbingly simple enough to implement in the generic case, with only one hiccup - `template std::ostream& operator<<(std::ostream &o, const std::vector& v) { std::copy(v.start(), v.end(), std::ostream_iterator(o, /* figure this part out and we're done */); return o; }` – Chris Lutz Jan 29 '11 at 01:32
@Chris Lutz: I've edited a corrected version of that into the answer, in case anybody wants it. At one time I wrote a more generic version that took various different container types, but I can't seem to find it at the moment... – Jerry Coffin Jan 29 '11 at 04:13

score 3 · Answer 2 · edited Jun 20 '13 at 02:14

3

use srand(time(0)) for seeding, this way you will really get a pseudorandom number

edited Jun 20 '13 at 02:14

pb2q

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answered Jan 29 '11 at 01:14

snoofkin

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score 2 · Answer 3 · answered Jan 29 '11 at 01:13

2

It looks like your program is only printing out one value:

cout << r << endl;

even though it looks like the given loop correctly generates the right number of random numbers. Are you sure that you're not creating the right number of numbers?

answered Jan 29 '11 at 01:13

templatetypedef

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score 0 · Answer 4 · edited Jun 20 '13 at 02:13

0

move cout << r << endl; inside your loop, then it will display the rand number and continue in its loop.

edited Jun 20 '13 at 02:13

pb2q

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answered Jun 20 '13 at 01:54

darkrogue83

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Creating Random Number Using Vectors in C++

4 Answers4