Can a map's key be shared with part of the value? - c++

Question

Can a std::map's or std::unordered_map's key be shared with part of the value? Especially if the key is non-trivial, say like a std::string?
As a simple example let's take a Person object:
struct Person {
// lots of other values
std::string name;
}
std::unordered_map<std::string, std::shared_ptr<Person>> people;
void insertPerson(std::shared_ptr<Person>& p) {
people[p.name] = p;
// ^^^^^^
// copy of name string
}
std::shared_ptr<Person> lookupPerson(const std::string& name) const {
return people[name];
}
My first thought is a wrapper around the name that points to the person, but I cannot figure out how to do a lookup by name.

Answer 1

For your purpose, a std::map can be considered a std::set containing std::pair's which is ordered (and thus efficiently accessible) according to the first element of the pair.
This view is particularly useful if key and value elements are partly identical, because then you do not need to artificially separate value and key elements for a set (and neither you need to write wrappers around the values which select the key).
Instead, one only has to provide a custom ordering function which works on the set and extracts the relevant key part.
Following this idea, your example becomes
auto set_order = [](auto const& p, auto const& s) { return p->name < s->name; };
std::set<std::shared_ptr<Person>, decltype(set_order)> people(set_order);
void insertPerson(std::shared_ptr<Person>& p) {
people.insert(p);
}
As an alternative, here you could also drop the custom comparison and order the set by the addresses in the shared pointer (which supports < and thus can be used directly in the set):
std::set<std::shared_ptr<Person> > people;
void insertPerson(std::shared_ptr<Person>& p) {
people.insert(p);
}
Replace set by unordered_set where needed (in general you then also need to provide a suitable hash function).
EDIT: The lookup can be performed using std:lower_bound:
std::shared_ptr<Person> lookupPerson(std::string const& s)
{
auto comp = [](auto const& p, auto const& s) { return p->name < s; };
return *std::lower_bound(std::begin(people), std::end(people), s, comp);
}
DEMO.
EDIT 2: However, given this more-or-less ugly stuff, you can also follow the lines of your primary idea and use a small wrapper around the value as key, something like
struct PersonKey
{
PersonKey(std::shared_ptr<Person> const& p) : s(p->name) {}
PersonKey(std::string const& _s) : s(_s) {}
std::string s;
bool operator<(PersonKey const& rhs) const
{
return s < rhs.s;
}
};
Use it like (untested)
std::map<PersonKey, std::shared_ptr<Person> > m;
auto sptr = std::make_shared<Person>("Peter");
m[PersonKey(sptr)]=sptr;
Lookup is done through
m[PersonKey("Peter")];
Now I like this better than my first suggestion ;-)

Answer 2

Here's an alternative to davidhigh's answer.
struct Person {
// lots of other values
std::string name;
}
struct StrPtrCmp {
bool operator()(const std::string* a, const std::string* b) const {
return *a < *b;
}
}
std::map<const std::string*, std::shared_ptr<Person>, StrPtrCmp> people();
void insertPerson(std::shared_ptr<Person>& p) {
people[&(p.name)] = p;
}
std::shared_ptr<Person> lookupPerson(const std::string& name) const {
return people[&name];
}
And a few edits to make it work with std::unordered_map:
struct StrPtrHash {
size_t operator()(const std::string* p) const {
return std::hash<std::string>()(*p);
}
};
struct StrPtrEquality {
bool operator()(const std::string* a, const std::string* b) const {
return std::equal_to<std::string>()(*a, *b);
}
};
std::unordered_map<const std::string*, std::shared_ptr<Person>, StrPtrHash, StrPtrEquality> people();