Because the set needs a comparison functor to work with. If you don't specify one, it will make a default-constructed one. In this case, since you're using a function-pointer type, the default-constructed one will be a null pointer, which can't be called; so instead, you have to provide the correct function pointer at run time.
A better approach might be to use a function class type (a.k.a. functor type); then the function call can be resolved at compile time, and a default-constructed object will do the right thing:
struct compare {
bool operator()(std::shared_ptr<myObject> &lhs,
std::shared_ptr<myObject> &rhs) const {
return lhs->value < rhs->value;
}
};
std::multiset<std::shared_ptr<myObject>, compare> myset;