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Generating a random double between a range of values

Im currently having trouble generating random numbers between -32.768 and 32.768. It keeps giving me the same values but with a small change in the decimal field. ex : 27.xxx.

Heres my code, any help would be appreciated.

#include <iostream>
#include <ctime>
#include <cstdlib>

using namespace std;

int main()
{
    srand( time(NULL) );
    double r = (68.556*rand()/RAND_MAX - 32.768);
    cout << r << endl;
    return 0;
}
like image 387
Olivier Avatar asked Feb 01 '13 02:02

Olivier


1 Answers

I should mention if you're using a C++11 compiler, you can use something like this, which is actually easier to read and harder to mess up:

#include <random>
#include <iostream>
#include <ctime>


int main()
{
    //Type of random number distribution
    std::uniform_real_distribution<double> dist(-32.768, 32.768);  //(min, max)

    //Mersenne Twister: Good quality random number generator
    std::mt19937 rng; 
    //Initialize with non-deterministic seeds
    rng.seed(std::random_device{}()); 

    // generate 10 random numbers.
    for (int i=0; i<10; i++)
    {
      std::cout << dist(rng) << std::endl;
    }
    return 0;
}

As bames53 pointed out, the above code can be made even shorter if you make full use of c++11:

#include <random>
#include <iostream>
#include <ctime>
#include <algorithm>
#include <iterator>

int main()
{
    std::mt19937 rng; 
    std::uniform_real_distribution<double> dist(-32.768, 32.768);  //(min, max)
    rng.seed(std::random_device{}()); //non-deterministic seed
    std::generate_n( 
         std::ostream_iterator<double>(std::cout, "\n"),
         10, 
         [&]{ return dist(rng);} ); 
    return 0;
}
like image 166
Carl Avatar answered Sep 20 '22 08:09

Carl