分配和内部状态？
分配和内部状态？
提问于 2018-04-28 12:20:15
在Stackoverflow中，有许多关于从未知范围生成均匀分布的整数的问题。例如
典型的解决方案是这样的：
inline std::mt19937 &engine()
{
thread_local std::mt19937 eng;
return eng;
}
int get_int_from_range(int from, int to)
{
std::uniform_int_distribution<int> dist(from, to);
return dist(engine());
}
下面的实现可靠吗？
int get_int_from_range(int from, int to)
{
using distribution_type = std::uniform_int_distribution<int>;
using param_type = typename distribution_type::param_type;
thread_local std::uniform_int_distribution<int> dist;
return dist(engine(), param_type(from, to));
}
Stack Overflow用户
发布于 2018-04-28 21:25:18
有趣的问题。
所以我想知道是否会通过不断重置它来干扰分配的工作方式（即在每次调用get_int_from_range时重新创建分配），我会得到正确分配的结果。
我已经编写了代码来测试这个uniform_int_distribution和poisson_distribution。如果你愿意，可以很容易地扩展它来测试另一个发行版。答案是肯定的。
代码：
#include <random>
#include <memory>
#include <chrono>
#include <utility>
typedef std::mt19937_64 engine_type;
inline size_t get_seed()
{ return std::chrono::system_clock::now().time_since_epoch().count(); }
engine_type& engine_singleton()
{
static std::unique_ptr<engine_type> ptr;
if ( !ptr )
ptr.reset( new engine_type(get_seed()) );
return *ptr;
}
// ------------------------------------------------------------------------
#include <cmath>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
void plot_distribution( const std::vector<double>& D, size_t mass = 200 )
{
const size_t n = D.size();
for ( size_t i = 0; i < n; ++i )
{
printf("%02ld: %s\n", i,
std::string(static_cast<size_t>(D[i]*mass),'*').c_str() );
}
}
double maximum_difference( const std::vector<double>& x, const std::vector<double>& y )
{
const size_t n = x.size();
double m = 0.0;
for ( size_t i = 0; i < n; ++i )
m = std::max( m, std::abs(x[i]-y[i]) );
return m;
}
实际测试代码：
#include <iostream>
#include <vector>
#include <cstdio>
#include <random>
#include <string>
#include <cmath>
void compare_uniform_distributions( int lo, int hi )
{
const size_t sample_size = 1e5;
// Initialize histograms
std::vector<double> H1( hi-lo+1, 0.0 ), H2( hi-lo+1, 0.0 );
// Initialize distribution
auto U = std::uniform_int_distribution<int>(lo,hi);
// Count!
for ( size_t i = 0; i < sample_size; ++i )
{
engine_type E(get_seed());
H1[ U(engine_singleton())-lo ] += 1.0;
H2[ U(E)-lo ] += 1.0;
}
// Normalize histograms to obtain "densities"
for ( size_t i = 0; i < H1.size(); ++i )
{
H1[i] /= sample_size;
H2[i] /= sample_size;
}
printf("Engine singleton:\n"); plot_distribution(H1);
printf("Engine creation :\n"); plot_distribution(H2);
printf("Maximum difference: %.3f\n", maximum_difference(H1,H2) );
std::cout<< std::string(50,'-') << std::endl << std::endl;
}
void compare_poisson_distributions( double mean )
{
const size_t sample_size = 1e5;
const size_t nbins = static_cast<size_t>(std::ceil(2*mean));
// Initialize histograms
std::vector<double> H1( nbins, 0.0 ), H2( nbins, 0.0 );
// Initialize distribution
auto U = std::poisson_distribution<int>(mean);
// Count!
for ( size_t i = 0; i < sample_size; ++i )
{
engine_type E(get_seed());
int u1 = U(engine_singleton());
int u2 = U(E);
if (u1 < nbins) H1[u1] += 1.0;
if (u2 < nbins) H2[u2] += 1.0;
}
// Normalize histograms to obtain "densities"
for ( size_t i = 0; i < H1.size(); ++i )
{
H1[i] /= sample_size;
H2[i] /= sample_size;
}
printf("Engine singleton:\n"); plot_distribution(H1);
printf("Engine creation :\n"); plot_distribution(H2);
printf("Maximum difference: %.3f\n", maximum_difference(H1,H2) );
std::cout<< std::string(50,'-') << std::endl << std::endl;
}
// ------------------------------------------------------------------------
int main()
{
compare_uniform_distributions( 0, 25 );
compare_poisson_distributions( 12 );
}
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/-100008312
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