Restrict random number generation to a range using <random> [duplicate]

Question

I'm generating random numbers in C++11. When I run my code

using namespace std;

static std::mt19937_64 rng;

int main() {
rng.seed(11);

    std::cout << rng() << std::endl;
}

It returns random numbers. However, I would like to restrict this to a range of numbers, say 1-1000. How would I do this, using the new <random> file/library introduced in C++11?

If you want them uniformly distributed, en.cppreference.com/w/cpp/numeric/random/…. — Qaz
– Qaz, Commented Jul 15, 2013 at 21:05
@andre Nope, the modulus operator doesn't generate a uniform distribution of numbers, check it out: cplusplus.com/reference/cstdlib/rand — user451498
– user451498, Commented Jul 15, 2013 at 21:06
Try this question -- noting that it's a different language, so not a dup (although those answers would do just fine). — sh1
– sh1, Commented Jul 15, 2013 at 21:06
% is strictly speaking not correct, to see that, assume the complete range was [0, 100) if you would % 80 random numbers from this range, the result would no longer be uniformly distributed. — Micha Wiedenmann
– Micha Wiedenmann, Commented Jul 15, 2013 at 21:07

Rapptz · Accepted Answer · 2013-07-15 21:06:54Z

17

Use std::uniform_int_distribution from <random>.

Example:

#include <random>
#include <iostream>

int main()
{
    std::random_device rd;
    std::mt19937 gen(rd());
    std::uniform_int_distribution<int> dis(1, 1000);

    std::cout << dis(gen) << '\n';
}

The <random> header includes a lot of other distributions that you can use. You can check it out here

answered Jul 15, 2013 at 21:06

Rapptz

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9 Comments

Qaz Over a year ago

+1 for easily accommodating other types of distributions, unlike the thinking process that went into my comment.

hichris123 Over a year ago

Wow, thanks the code and everything!

hichris123 Over a year ago

Also, how would I store this random number as a variable?

Rapptz Over a year ago

@user2581872 I don't get any errors. You might want to make that a separate question.

Keith Thompson Over a year ago

@user2581872: Call it multiple times: foo = dis(gen); bar = dis(gen);

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