std::sample() 具有整数范围

Question

如何在任何给定的整数范围内随机选取多个值？
对于std::sample()，
一种可能的实现是：

void Sample(
    int first, int last, std::vector<int> *out, std::size_t n, std::mt19937 *g)
{
    std::vector<int> in{};
    for (int i = first; i < last; ++i)
    {
        in.emplace_back(i);
    }

    std::sample(in.begin(), in.end(), std::back_inserter(*out), n, *g);
}

但我很犹豫，因为创建in向量似乎多余且不切实际，
特别是当first是 0 并且last是非常大的数字时。
是否有任何公共实现或库可以满足此要求？例如：

• sample(int first, int last, ...)或者，
• ranges::sample(int n, ...)

Answer 1

In C++20 you can sample on iota_view

#include <ranges>
#include <vector>
#include <algorithm>
#include <random>

void Sample(
  int first, int last, std::vector<int> *out, std::size_t n, std::mt19937 *g) 
{
  std::ranges::sample(
    std::views::iota(first, last), std::back_inserter(*out), n, *g);  
}

Demo

Note that the above does not work in libstdc++ because ranges::sample incorrectly uses the std::sample's implementation. Since iota_view's iterator only satisfies Cpp17InputIterator, std::sample requires that the output range must be a random access iterator, which is not the case for back_inserter_iterator.

The workaround is to pre-resize the vector.

void Sample(
  int first, int last, std::vector<int> *out, std::size_t n, std::mt19937 *g) 
{
  out->resize(n);
  std::ranges::sample(
    std::views::iota(first, last), out->begin(), n, *g);  
}

which works in both libstdc++ and MSVC-STL.