What is the difference between unordered_map::emplace and unordered_map::insert in C++?

Harsh M. Shah picture Harsh M. Shah · Oct 19, 2014 · Viewed 26.5k times · Source

What is the difference between std::unordered_map::emplace and std::unordered_map::insert in C++?

Answer

Chris Drew picture Chris Drew · Oct 19, 2014

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
}