Why does rand() yield the same sequence of numbers on every run?

Question

Every time I run a program with rand() it gives me the same results.

Example:

#include <iostream>
#include <cstdlib>

using namespace std;

int random (int low, int high) {
    if (low > high)
        return high;
    return low + (rand() % (high - low + 1));
}

int main (int argc, char* argv []) {
    for (int i = 0; i < 5; i++)
        cout << random (2, 5) << endl;
}

Output:

3
5
4
2
3

Each time I run the program it outputs the same numbers every time. Is there a way around this?

Answer 1

The seed for the random number generator is not set.

If you call srand((unsigned int)time(NULL)) then you will get more random results:

#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;

int main() {
    srand((unsigned int)time(NULL));
    cout << rand() << endl;
    return 0;
}

The reason is that a random number generated from the rand() function isn't actually random. It simply is a transformation. Wikipedia gives a better explanation of the meaning of pseudorandom number generator: deterministic random bit generator. Every time you call rand() it takes the seed and/or the last random number(s) generated (the C standard doesn't specify the algorithm used, though C++11 has facilities for specifying some popular algorithms), runs a mathematical operation on those numbers, and returns the result. So if the seed state is the same each time (as it is if you don't call srand with a truly random number), then you will always get the same 'random' numbers out.

If you want to know more, you can read the following:

http://www.dreamincode.net/forums/topic/24225-random-number-generation-102/

http://www.dreamincode.net/forums/topic/29294-making-pseudo-random-number-generators-more-random/