问题
I know I could use the following:
template <typename Pair>
struct ComparePairThroughSecond : public std::unary_function<Pair, bool>
{
bool operator ()(const Pair& p1, const Pair& p2) const
{
return p1.second < p2.second;
}
};
std::set<std::pair<int, long>, ComparePairThroughSecond> somevar;
but wondered if it could be done with boost::bind
回答1:
How about the following one. I'm using boost::function to 'erase' the actual type of the comparator. The comparator is created using boost:bind itself.
typedef std::pair<int, int> IntPair;
typedef boost::function<bool (const IntPair &, const IntPair &)> Comparator;
Comparator c = boost::bind(&IntPair::second, _1) < boost::bind(&IntPair::second, _2);
std::set<IntPair, Comparator> s(c);
s.insert(IntPair(5,6));
s.insert(IntPair(3,4));
s.insert(IntPair(1,2));
BOOST_FOREACH(IntPair const & p, s)
{
std::cout << p.second;
}
回答2:
The problem is that -- unless you write your code as a template or use C++0x features -- you have to name the type of the boost::bind expression. But those types usually have very complicated names.
Template argument deduction in C++98:
template<class Fun>
void main_main(Fun fun) {
set<pair<int,long>,Fun> s (fun);
…
}
int main() {
main_main(…boost::bind(…)…);
}
With auto and decltype in C++0x:
int main() {
auto fun = …boost::bind(…)…;
set<pair<int,long>,decltype(fun)> s (fun);
main_main(boost::bind(…));
}
As for the actual bind expression, I think it's something like this:
typedef std::pair<int,long> pil;
boost::bind(&pil::second,_1) < boost::bind(&pil::second,_2)
(untested)
来源:https://stackoverflow.com/questions/2958226/how-do-i-create-a-set-with-stdpair-thats-sorted-based-on-the-second-pair-mem