Generating a random double between a range of values

问题

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;
}

回答1:

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;
}

回答2:

Also, If you are not using c++ 11 you can use the following function instead:

double randDouble(double precision, double lowerBound, double upperBound) {
 double random;
 random = static_cast<double>(((rand()%(static_cast<int>(std::pow(10,precision)*(upperBound - lowerBound) + 1))) + lowerBound*std::pow(10,precision)))/std::pow(10,precision);
 return random;
}

回答3:

So, I think this is a typical case of "using time(NULL) isn't a great way of seeding random numbers for runs that start close together". There isn't that many bits that change in time(NULL) from one call to the next, so random numbers are fairly similar. This is not a new phenomena - if you google "my random numbers aren't very random", you'll find LOTS of this.

There are a few different solutions - getting a microsecond or nanosecond time would be the simplest choice - in Linux gettimeofday will give you a microsecond time as part of the struct.

回答4:

It seams to be plainly obvious but some of the examples say otherwise... but i thought when you divide 1 int with another you always get an int? and you need to type cast each int to double/float before you divide them.

ie: double r = (68.556* (double)rand()/(double)RAND_MAX - 32.768);

also if you call srand() every time you call rand() you reset the seed which results in similar values returned every time instead of ''random'' ones.

回答5:

I've added a for loop to your program:

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

using namespace std;

int main () {
    srand(time (NULL));

    for (int i = 0; i < 10; ++i) {
        double r = ((68.556 * rand () / RAND_MAX) - 32.768);

        cout << r << endl;
    }

    return 0;
}

Example output:

It seems Okay to me. I've added the code on Ideone for you.

Here are four runs:

Perhaps you're not waiting at least a second between runs?

来源：https://stackoverflow.com/questions/14638739/generating-a-random-double-between-a-range-of-values

标签

c++

algorithm

random

prng

genetic