C++:为什么这个方法每次都返回相同的随机数？

Question

我有一个类，其中有一些数据存储在一个 vector 中，还有一个方法应该选择一个随机元素并返回它，但是当我运行它时，它每次都返回相同的元素。

这是一个基于我的代码的简化示例:

#include <iostream>
#include <random>
#include <vector>

class MyObj{
  private:
    std::vector<int> set_data;

  public:
    MyObj(int num_elements){
      for (int i = 0; i < num_elements; ++i){
        set_data.push_back(i); // just so that there is some data in there
      }
    };
    int getRandomElement(std::mt19937 rng){
      std::uniform_int_distribution<int> uni(0,set_data.size()-1);
      int idx = uni(rng);
      return set_data[idx];
    };
};


int main()
{
  std::random_device r;
  std::seed_seq seed{r(), r(), r(), r(), r(), r(), r(), r()};
  std::mt19937 rng = std::mt19937(seed);

  MyObj temp(50);

  for (int i = 0; i < 20; i++){
    std::cout << "getting random element: " << temp.getRandomElement(rng) << std::endl;
  }
}

输出是:

getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19
getting random element: 19

有什么明显的地方我做错了吗？

Answer 1

int getRandomElement(std::mt19937 rng){
  std::uniform_int_distribution<int> uni(0,set_data.size()-1);
  int idx = uni(rng);
  return set_data[idx];
};

这是按值传递，这意味着 RNG 的状态是从原始状态复制的，而原始状态永远不会更新。将此方法更改为通过引用传递应该可以解决此问题。

int getRandomElement(std::mt19937 & rng){