Only instance of std::mt19937 repeats values in c++11

Question

In the program, often in different classes are generated random numbers. So I want to create a class that returns a single instance of the generator std::mt19937. I also take into account that some compilers do not work with std::random_device (To do this, check the value of entropy). I have created a class singleton.

#include <iostream>
#include <random>
#include <chrono>

class RandomGenerator
{
public:
    static RandomGenerator& Instance() {
        static RandomGenerator s;
        return s;
    }
    std::mt19937 get();

private:
    RandomGenerator();
    ~RandomGenerator() {}

    RandomGenerator(RandomGenerator const&) = delete;
    RandomGenerator& operator= (RandomGenerator const&) = delete;

    std::mt19937 mt;
};

RandomGenerator::RandomGenerator() {
    std::random_device rd;

    if (rd.entropy() != 0) {
        mt.seed(rd());
    }
    else {
        auto seed = std::chrono::high_resolution_clock::now().time_since_epoch().count();
        mt.seed(seed);
    }
}

std::mt19937 RandomGenerator::get() {
    return mt;
}

int main() {

    std::mt19937 &mt = RandomGenerator::Instance().get();
    std::uniform_real_distribution<double> dist(0.0, 1.0);
    for (std::size_t i = 0; i < 5; i++)
        std::cout << dist(mt) << "\n";

    std::cout << "\n";

    std::mt19937 &mt2 = RandomGenerator::Instance().get();
    std::uniform_real_distribution<double> dist2(0.0, 1.0);
    for (std::size_t i = 0; i < 5; i++)
        std::cout << dist2(mt2) << "\n";

    return 0;
}

But when I get out of class generator std::mt19937, random numbers begin to repeat. How to avoid it?

P.S. Is there a better way to initialize the generator than time?

Solution

Tested this under the following compilers: Visual Studio, MinGW, DevC++.

#include <iostream>
#include <random>
#include <chrono>

class RandomGenerator
{
public:
    static RandomGenerator& Instance() {
        static RandomGenerator s;
        return s;
    }
    std::mt19937 & get();

private:
    RandomGenerator();
    ~RandomGenerator() {}

    RandomGenerator(RandomGenerator const&) = delete;
    RandomGenerator& operator= (RandomGenerator const&) = delete;

    std::mt19937 mt;
};

RandomGenerator::RandomGenerator() {
    std::random_device rd;

    if (rd.entropy() != 0) {
        mt.seed(rd());
    }
    else {
        auto seed = std::chrono::high_resolution_clock::now().time_since_epoch().count();
        mt.seed(seed);
    }
}

std::mt19937 & RandomGenerator::get() {
    return mt;
}

int main() {

    std::mt19937 &mt = RandomGenerator::Instance().get();
    std::uniform_real_distribution<double> dist(0.0, 1.0);
    for (std::size_t i = 0; i < 5; i++)
        std::cout << dist(mt) << "\n";

    std::cout << "\n";

    std::mt19937 &mt2 = RandomGenerator::Instance().get();
    std::uniform_real_distribution<double> dist2(0.0, 1.0);
    for (std::size_t i = 0; i < 5; i++)
        std::cout << dist2(mt2) << "\n";

    return 0;
}

Youre returning it by value, which means any internal state modification you make by generating random numbers is lost, thus repeating the patter every time you call get() again. — Borgleader, May 10 '17 at 18:53
`std::mt19937 get();` returns by value. This code wouldn't compile with a standards compliant compiler: `std::mt19937 &mt = RandomGenerator::Instance().get();`. — juanchopanza, May 10 '17 at 18:53
[Generating random numbers](http://dilbert.com/strip/2001-10-25) — Wade Tyler, May 10 '17 at 18:54
This will not compile. `&mt = RandomGenerator::Instance().get();` — gsamaras, May 10 '17 at 18:55
@gsamaras That statement is non-standard, but it compiles in Visual Studio as it's part of the language extensions it likes to enable by default. — François Andrieux, May 10 '17 at 18:57
@gsamaras Why this will not compile? Visual Studio compiler and MinGW compiler quietly compiles. — Harrix, May 10 '17 at 19:02
Because `error: non-const lvalue reference to type 'mersenne_twister_engine<...>' cannot bind to a temporary of type 'mersenne_twister_engine<...>'`. That's what you get when working with non-standard compilers :/ — gsamaras, May 10 '17 at 19:09
@gsamaras What compiler are you using? On DevС++ also compiled. — Harrix, May 10 '17 at 19:18
@Harrix it works on Visual Studio as an extension. In order to catch subtle non-standard behaviours like this, I know that Visual Studio has a flag for it (might only be on 2017, but it's probably on 2015 too). I don't remember the flag, however. — Justin, May 10 '17 at 19:22
@Justin What kind of code are you talking about? Code with the correct method you checked out the visual Studio 2015 compiler mingw (QT version 5.8) and devc++. — Harrix, May 10 '17 at 19:29
@gsamaras True. In gcc bad code do not compile. The corrected version is compiled. — Harrix, May 11 '17 at 04:20
Found the compiler flag I was talking about; it's in VS 2017: [`/permissive-`](https://blogs.msdn.microsoft.com/vcblog/2016/11/16/permissive-switch/) — Justin, May 11 '17 at 06:10

score 5 · Accepted Answer · answered May 10 '17 at 18:54

std::mt19937 get(); returns a copy. Every time you call get() you make a copy of the initial state of the engine. mt19937 is a pseudo-random engine, each state produces a predetermined sequence. If the state of two instances are identical, they will produce the same sequence. Make the function return a reference so that the state of the singleton instance is updated with each new number generated.

std::mt19937 & RandomGenerator::get() {
    return mt;
}

Only instance of std::mt19937 repeats values in c++11

1 Answers1