c++ - C++11 中的通用 STL 兼容直方图

Question

这是我第一次尝试 C++ 中的通用直方图模板函数，并使用 GCC 4.6 进行测试。但是，我想合并dense_histogram()和sparse_histogram()到一个通用的通用函数模板中。问题在于特定于密集的构造函数 H h(n, 0)在稀疏版本 H h 中既没有定义也没有相关性。有没有办法以某种巧妙的 C++ 常规方式或静态地通过 Boost Type.Traits ( #include <boost/type_traits.hpp> ) 使用条件编译来解决这个问题？

#include <algorithm>
#include <limits>
#include <algorithm>
#include <vector>
#include <unordered_map>

namespace std
{

/*!
 * \em Dense Histogram of \p a.
 *
 * \tparam V is Value Type.
 * \tparam C is Count (Bin) Type.
 * \tparam H is Histogram Storage Type, typically a vector.
 *
 * \param[in] x is a set of the input data set
 */
template <class V, class C = size_t, class H = vector<C> >
inline
H dense_histogram(const V & x)
{
    typedef typename V::value_type E; // element type
    size_t n = (static_cast<C>(1)) << (8*sizeof(E)); // maximum number of possible elements for dense variant
    H h(n, 0);                       // histogram
    C bmax = 0;                      // bin max
    for_each(begin(x), end(x),  // C++11
             [&h, &bmax] (const E & e) { // value element
                 h[e]++;
                 bmax = std::max(bmax, h[e]);
             });
    return h;
}
template <class V, class H = vector<size_t> > H make_dense_histogram(const V & x) { return dense_histogram<V, size_t, H>(x); }

/*!
 * \em Sparse Histogram of \p a.
 *
 * \tparam V is Value Type.
 * \tparam C is Count (Bin) Type.
 * \tparam H is Histogram Structure Type, typically a unordered_map.
 *
 * \param[in] x is a set of the input data set
 */
template <class V, class C = size_t, class H = unordered_map<typename V::value_type, C> >
inline
H sparse_histogram(const V & x)
{
    typedef typename V::value_type E; // element type
    H h;                        // histogram
    C bmax = 0;                 // bin max
    for_each(begin(x), end(x), // C++11
             [&h,&bmax] (const E & e) { // value element
                 h[e]++;
                 bmax = std::max(bmax, h[e]);
             });
    return h;
}
template <class V, class H = unordered_map<typename V::value_type, size_t> > H make_sparse_histogram(const V & x) { return sparse_histogram<V, size_t, H>(x); }

}

运行使用

Answer 1

我认为你应该简单地将公共(public)部分放在第三个函数中，留下dense_histogram和sparse_histogram来创建h并调用该实现函数:

template <class V, class C = size_t, class H>
inline void histogram_impl(const V & x, H& h) {
    typedef typename V::value_type E; // element type
    C bmax = 0;                      // bin max
    for_each(begin(x), end(x),  // C++11
             [&h, &bmax] (const E & e) { // value element
                 h[e]++;
                 bmax = std::max(bmax, h[e]);
             });
    return h;
}
template <class V, class C = size_t, class H = vector<C> >
inline H dense_histogram(const V & x) {
    typedef typename V::value_type E; // element type
    size_t n = (static_cast<C>(1)) << (8*sizeof(E)); // maximum number of possible elements for dense variant
    H h(n, 0);                       // histogram
    histogram_impl(x, h);
    return h;
}
template <class V, class C = size_t, class H = unordered_map<typename V::value_type, C> >
inline H sparse_histogram(const V & x) {
    H h;                        // histogram
    histogram_impl(x, h);
    return h;
}

但是，既然您要求它:当您正在处理容器时，我会假设它们有一个便宜的移动，因此您可以定义一个创建特征来生成容器并将其移动到本地变量中。然后您可以编写自己的适当构造函数检测，如下所示:

template<typename T> struct has_explicit_length_constructor{
private:
   template<typename U>
   decltype(U(0, 0), void(), std::true_type()) test(int x);
   template<typename>
   std::false_type test(...);
  typedef decltype(test<T>(0)) constant_type;
public:
   constexpr bool value = constant_type::value;
};

template<class H, bool B = has_explicit_length_constructor<H>::value> struct histogram_creation_trait;
template<class H> struct histogram_creation_trait<H, true> {
  static H create()  {
    size_t n = (static_cast<C>(1)) << (8*sizeof(typename V::value_type));
    return H(n, 0);  
  }
};
template<class H> struct histogram_creation_trait<H, false>
{ static H create()  { return H(); } };

template <class V, class C = size_t, class Ht>
inline void histogram_impl(const V & x, H& h, Trait) {
    typedef typename V::value_type E; // element type
    C bmax = 0;                      // bin max
    H h = histogram_creation_trait<H>::create();
    for_each(begin(x), end(x),  // C++11
             [&h, &bmax] (const E & e) { // value element
                 h[e]++;
                 bmax = std::max(bmax, h[e]);
             });
    return h;
}
template <class V, class H = vector<size_t> > H make_dense_histogram(const V & x) { return histogram_impl<V, size_t, H>(x); }
template <class V, class H = unordered_map<typename V::value_type, size_t> > H make_sparse_histogram(const V & x) { return histogram_impl<V, size_t, H>(x); }

顺便说一句:按照标准，将您自己的方法添加到 std 是 UB ([namespace.std] $17.6.4.2.1 p1):

The behavior of a C++ program is undefined if it adds declarations or definitions to namespace std or to a namespace within namespace std unless otherwise specified. A program may add a template specialization for any standard library template to namespace std only if the declaration depends on a user-defined type and the specialization meets the standard library requirements for the original template and is not explicitly prohibited.