Is C++ bimap possible with one side of view having different key than other side of the view value? How to do that?

问题

At the beginning I needed a map, so I used std::map.
Then, some requirements were added and I needed to also get "keys" for "value" (foos for bar), so I used

boost::bimaps::bimap<
  boost::bimaps::unordered_set_of<boost::bimaps::tagged<std::string, foo>>, 
  boost::bimaps::multiset_of<boost::bimaps::tagged<std::string, bar>>>

And after that, some more requirements were added, so now I need to store a number for every foo and from the right side view I need to be able to call <bimap>.righ.find(bar) and get pairs of (foo + number stored for foo), but I still want to be able to call <bimap>.left.find(foo) and get bar.

How to achieve that? I would prefer some modern C++ over boost if possible, but I guess it is harder to have bimap functionality without the boost.

EDIT: I should note that size matters, so I don't want to store any part involved twice and the speed also matters.

I should have something like
"foo1"+100 <-> "bar1" and "foo2"+300 <-> "bar4".
and I want to be able to call <bimap>.left.find("foo1") and get "bar1",
but also <bimap>.right.find("bar1") and get pair("foo1", 100).

回答1:

#include <boost/multi_index/hashed_index.hpp>
#include <boost/bimap/bimap.hpp>

using namespace std;

struct ElementType { 
  string foo; 
  string bar;
  uint64_t number; 
};

using namespace boost::multi_index;

using my_bimap = multi_index_container<
  ElementType,
  indexed_by<
    hashed_unique<member<ElementType, string, &ElementType::foo>>,
    ordered_non_unique<member<ElementType, string, &ElementType::bar>>
  >
>;

int main() {
  my_bimap instance;

  instance.insert({"foo", "bar", 0});
  instance.insert({"bar", "bar", 1});

  cout << instance.get<0>().find("bar")->foo << endl;
  cout << instance.get<0>().find("bar")->bar << endl;
  cout << instance.get<0>().find("bar")->number << endl;
  auto range = instance.get<1>().equal_range("bar");
  for (auto it = range.first; it != range.second; ++it) {
    cout << it->foo << endl;
    cout << it->number << endl;
  }

  cin.sync();
  cin.ignore();
}

Output:

bar
bar
1
foo
0
bar
1

来源：https://stackoverflow.com/questions/50684467/is-c-bimap-possible-with-one-side-of-view-having-different-key-than-other-side