[Boost-users] [multi_index] do erase() call key extractors?

23 Dec 2015

      This is probably rather obvious, but I wasn't able to locate this guarantee
in MIC docs. Take,

struct Foo {
    explicit Foo(Bar* b = nullptr) : bar(b) {}
    Bar* bar;
};

int getBarValue (const Foo& f) {
     assert( f.bar );
     return f.bar->getValue();
}

boost::multi_index< Foo,  indexed_by<
<  hashed_unique < member >>
,   hashed_non_unique < global_fun>
...
...
FooIndex;
void grill() {
    FooIndex  fooIndex;

    Bar* b;
    {
         auto bWork = make_unique<Bar>();
         // ...
         fooIndex.emplace( bWork.get() );
         b = bWork.get();
     }

     fooIndex.erase(b);

     // ...
}

Please ignore the fact that raw-pointers are potentially dangerous etc. The
question is does erase() call getBarValue() despite the fact I am using the
first index? More generally, do key extractors get called *at all* except
at the point of insertion?

Thanks,
Nick

[Boost-users] [multi_index] do erase() call key extractors?

Nick Stokes