boost multi_index提取的密钥是否被缓存？

Question

我正在使用boost :: multi_index,我希望根据数据类型对其进行索引.但是,此数据类型的size()成员函数执行起来很昂贵.multi_index是否缓存从其关键提取器获取的值？

例如,如果我创建了一个带有成员函数键(element.size())的有序索引的multi_index容器,并插入了一个大小将其放在容器中间某个位置的元素,那么容器是否会重新调用该大小()成员函数在查找正确的插入点之前遍历其内部数据结构时访问的所有元素？

Answer 1

好吧,成员函数索引器的文档说它们调用引用的成员函数:http://www.boost.org/doc/libs/1_46_0/libs/multi_index/doc/reference/key_extraction.html#key_extractors

但如有疑问,请查看:

#include <boost/multi_index_container.hpp>
#include <boost/multi_index/mem_fun.hpp>
#include <boost/multi_index/indexed_by.hpp>
#include <boost/multi_index/hashed_index.hpp>
#include <boost/multi_index/ordered_index.hpp>
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

namespace bmi = boost::multi_index;

int g_calls = 0;
struct A
{
  explicit A(int sz) : m_size(sz) { }
  int size() const { ++g_calls; return m_size; }
private:
  int m_size;
};

typedef boost::multi_index_container<
  A*,
  bmi::indexed_by<
    bmi::ordered_non_unique<
      BOOST_MULTI_INDEX_CONST_MEM_FUN(A,int,A::size)
    >
  >
> container_t;

int main()
{
  container_t cont;
  int n = 100;
  vector<int> o(2*n+1);
  for( int i = 0; i != 2*n+1; ++i )
    o[i] = i;
  random_shuffle(o.begin(), o.end());

  for( int i = 0; i != n; ++i )
    cont.insert(new A(o[i]));
  cout << "Calls to A::size(): "<< g_calls << endl;
  for( int i = n+1; i <= 2*n; ++i )
    cont.insert(new A(o[i]));
  cout << "Calls to A::size(): "<< g_calls << endl;
  cont.insert(new A(o[n]));
  cout << "Calls to A::size(): "<< g_calls << endl;
  for( int i = 0; i != o.size(); ++i )
    cont.find(o[i]);
  cout << "Calls after calling find " << o.size() << " times: "<< g_calls << endl;
  return 0;
}

给出以下输出(使用Boost 1.46):

Calls to A::size(): 629
Calls to A::size(): 1465
Calls to A::size(): 1474
Calls after calling find 201 times: 3262

所以,似乎答案是否定的,它不会缓存值.