What is the difference between unordered_map::emplace and unordered_map::insert in C++?
What is the difference between std::unordered_map::emplace
and std::unordered_map::insert
in C++?
Solution 1:
unordered_map::insert
copies or moves a key-value pair into the container. It is overloaded to accept reference-to-const or an rvalue reference:
std::pair<iterator,bool> insert(const std::pair<const Key, T>& value);
template<class P>
std::pair<iterator,bool> insert(P&& value);
unordered_map::emplace
allows you to avoid unnecessary copies or moves by constructing the element in place. It uses perfect forwarding and a variadic template to forward arguments to the constructor of the key-value pair:
template<class... Args>
std::pair<iterator,bool> emplace(Args&&... args);
But there is a great deal of overlap between the two functions. emplace
can be used to forward to the copy/move constructor of the key-value pair which allows it to be used just as insert
would. This means that use of emplace
doesn't guarantee you will avoid copies or moves. Also the version of insert
that takes an rvalue-reference is actually templated and accepts any type P
such that the key-value pair is constructible from P
.
Scott Meyers says:
In principle, emplacement functions should sometimes be more efficient than their insertion counterparts, and they should never be less efficient.
( Edit: Howard Hinnant ran some experiments that showed sometimes insert
is faster than emplace
)
If you definitely do want to copy/move into the container it might be wise to use insert
because you are more likely to get a compilation error if you pass incorrect arguments. You need to be more careful you are passing the correct arguments to the emplacement functions.
Most implementations of unordered_map::emplace
will cause memory to be dynamically allocated for the new pair even if the map contains an item with that key already and the emplace
will fail. This means that if there is a good chance that an emplace
will fail you may get better performance using insert to avoid unneccessary dynamic memory allocations.
Small example:
#include <unordered_map>
#include <iostream>
int main() {
auto employee1 = std::pair<int, std::string>{1, "John Smith"};
auto employees = std::unordered_map<int, std::string>{};
employees.insert(employee1); // copy insertion
employees.insert(std::make_pair(2, "Mary Jones")); // move insertion
employees.emplace(3, "James Brown"); // construct in-place
for (const auto& employee : employees)
std::cout << employee.first << ": " << employee.second << "\n";
}
Edit2: On request. It is also possible to use unordered_map::emplace
with a key or value that takes more than one constructor parameter. Using the std::pair
piecewise constructor you can still avoid unnecessary copies or moves.
#include <unordered_map>
#include <iostream>
struct Employee {
std::string firstname;
std::string lastname;
Employee(const std::string& firstname, const std::string& lastname)
: firstname(firstname), lastname(lastname){}
};
int main() {
auto employees = std::unordered_map<int, Employee>{};
auto employee1 = std::pair<int, Employee>{1, Employee{"John", "Smith"}};
employees.insert(employee1); // copy insertion
employees.insert(std::make_pair(2, Employee{"Mary", "Jones"})); // move insertion
employees.emplace(3, Employee("Sam", "Thomas")); // emplace with pre-constructed Employee
employees.emplace(std::piecewise_construct,
std::forward_as_tuple(4),
std::forward_as_tuple("James", "Brown")); // construct in-place
}
Solution 2:
The difference between emplace()
and insert()
has already been well explained in Chris Drew's answer. However, for the sake of completeness I'd like to add that since C++17 std::unordered_map
provides two new insertion methods: try_emplace()
and insert_or_assign()
. Let me summarize these methods briefly:
-
try_emplace()
is an "improved" version ofemplace()
. In contrast toemplace()
,try_emplace()
doesn't modify its arguments (due to move operations) if insertion fails due to a key already existing in theunordered_map
. -
insert_or_assign()
is an "improved" version ofoperator[]
. In contrast tooperator[]
,insert_or_assign()
doesn't require the value type of theunordered_map
to be default constructible.
I have written a more detailed answer on the above mentioned new insertion methods for std::map
here. That answer also applies to std::unordered_map
.
Simple example code on Coliru