C++ shared_ptr equality operator

Question

The equality operator for shared_ptr's is defined as follows:

template<class T, class U> inline bool operator==(
    shared_ptr<T> const & a, shared_ptr<U> const & b)
{
    return a.get() == b.get();
}

This seems broken. Would it not have been better to forward the equality to what a and b are pointing to? Or would that be an unfair restriction on users of the library (in that they have to provide an equality operator) ?

If I have a map or a hash_table containing shared_ptrs, then the current definition makes equality unusable. For example, consider

std::map<int, std::tr1::shared_ptr<T> > m1, m2;

Won't we want to check that the ptrs for each int in m1 and m2 are pointing to the same value ?

I can implement my own equality by flattening m1, m2 out (constructing sets from each, dereferencing shared_ptrs along the way). Is there an STL trick that will accomplish this or some other way to test equality in the presence of shared_ptrs neatly ?

Answer 1

It's not broken, because a shared_ptr is conceptually a pointer, therefore it implements pointer-wise equality. When you test two pointers for equality, you want to know whether they point to the same place in memory.