Zitat von Howard Hinnant <howard.hinnant@gmail.com>:
At each remove() the node_ptr is ignored/destructed, thus deallocating the node. There is no iterator increment.
This is a performance-tested solution with your test program:
20000: 0m0.012s // your time is 0m0.244s 40000: 0m0.017s // your time is 0m0.940s 60000: 0m0.024s // your time is 0m5.584s 80000: 0m0.028s // your time is 0m4.232s 100000: 0m0.037s // your time is 0m34.814s 120000: 0m0.041s // your time is 0m33.486s
The bad news is that the LWG will consider this much too inventive at this late date. That being said, I believe it solves several real world performance problems (this one and others listed in LWG 839). I offer it here in case someone wants to implement it for boost containers.
In case you don't like the "too inventive" response from the C++ committee, I'm the one to yell at. I will happily forward your response to the committee.
I think none of the solutions are perfect, with Option 1(void result) being the one that makes the most sense to me. as others have noted it wouldn't be the first time a container violates that requirement, and a returned iterator is of very limited use in an unordered container anyway. but while we're being inventive, I would like to see the following considered, for the long-term. if I am not mistaken we're dealing with a function overloading problem here. overloading by return type. template<class T,...> class unordered_set{ ... void erase(iterator); // (1) iterator erase(iterator); // (2) ... }; s.erase(it); //calls (1) it=s.erase(it); //calls (2) this obviously requires language support, but it would solve the problem beyond this specific case. problems like this are quite common, and are usually (when you're not bound to a Container concept) solved like this: void erase(iterator in,iterator *out=0); or void erase(iterator in); void erase(iterator in,iterator &out); you can find signatures like this in several boost libraries. more examples of overloading by return type: int f(); // (1) float f(); // (2) void f(); // (3) int a=f(); //calls (1) f(); //calls (3) (float)f(); //calls (2) void g(int); void g(float); g(f()); //ambiguous Regards, Stefan