insert vs emplace vs operator[] in c++ map

Question

I\'m using maps for the first time and I realized that there are many ways to insert an element. You can use emplace(), operator[] or insert(

Answer 1

In the particular case of a map the old options were only two: operator[] and insert (different flavors of insert). So I will start explaining those.

The operator[] is a find-or-add operator. It will try to find an element with the given key inside the map, and if it exists it will return a reference to the stored value. If it does not, it will create a new element inserted in place with default initialization and return a reference to it.

The insert function (in the single element flavor) takes a value_type (std::pair<const Key,Value>), it uses the key (first member) and tries to insert it. Because std::map does not allow for duplicates if there is an existing element it will not insert anything.

The first difference between the two is that operator[] needs to be able to construct a default initialized value, and it is thus unusable for value types that cannot be default initialized. The second difference between the two is what happens when there is already an element with the given key. The insert function will not modify the state of the map, but instead return an iterator to the element (and a false indicating that it was not inserted).

// assume m is std::map<int,int> already has an element with key 5 and value 0
m[5] = 10;                      // postcondition: m[5] == 10
m.insert(std::make_pair(5,15)); // m[5] is still 10

In the case of insert the argument is an object of value_type, which can be created in different ways. You can directly construct it with the appropriate type or pass any object from which the value_type can be constructed, which is where std::make_pair comes into play, as it allows for simple creation of std::pair objects, although it is probably not what you want...

The net effect of the following calls is similar:

K t; V u;
std::map<K,V> m;           // std::map<K,V>::value_type is std::pair<const K,V>

m.insert( std::pair<const K,V>(t,u) );      // 1
m.insert( std::map<K,V>::value_type(t,u) ); // 2
m.insert( std::make_pair(t,u) );            // 3

But the are not really the same... [1] and [2] are actually equivalent. In both cases the code creates a temporary object of the same type (std::pair<const K,V>) and passes it to the insert function. The insert function will create the appropriate node in the binary search tree and then copy the value_type part from the argument to the node. The advantage of using value_type is that, well, value_type always matches value_type, you cannot mistype the type of the std::pair arguments!

The difference is in [3]. The function std::make_pair is a template function that will create a std::pair. The signature is:

template <typename T, typename U>
std::pair<T,U> make_pair(T const & t, U const & u );

I have intentionally not provided the template arguments to std::make_pair, as that is the common usage. And the implication is that the template arguments are deduced from the call, in this case to be T==K,U==V, so the call to std::make_pair will return a std::pair<K,V> (note the missing const). The signature requires value_type that is close but not the same as the returned value from the call to std::make_pair. Because it is close enough it will create a temporary of the correct type and copy initialize it. That will in turn be copied to the node, creating a total of two copies.

This can be fixed by providing the template arguments:

m.insert( std::make_pair<const K,V>(t,u) );  // 4

But that is still error prone in the same way that explicitly typing the type in case [1].

Up to this point, we have different ways of calling insert that require the creation of the value_type externally and the copy of that object into the container. Alternatively you can use operator[] if the type is default constructible and assignable (intentionally focusing only in m[k]=v), and it requires the default initialization of one object and the copy of the value into that object.

In C++11, with variadic templates and perfect forwarding there is a new way of adding elements into a container by means of emplacing (creating in place). The emplace functions in the different containers do basically the same thing: instead of getting a source from which to copy into the container, the function takes the parameters that will be forwarded to the constructor of the object stored in the container.

m.emplace(t,u);               // 5

In [5], the std::pair<const K, V> is not created and passed to emplace, but rather references to the t and u object are passed to emplace that forwards them to the constructor of the value_type subobject inside the data structure. In this case no copies of the std::pair<const K,V> are done at all, which is the advantage of emplace over the C++03 alternatives. As in the case of insert it will not override the value in the map.

---

An interesting question that I had not thought about is how emplace can actually be implemented for a map, and that is not a simple problem in the general case.

Answer 2

Emplace: Takes advantage of the rvalue reference to use the actual objects that you have already created. This means that no copy or move constructor is called, good for LARGE objects! O(log(N)) time.

Insert: Has overloads for standard lvalue reference and rvalue reference, as well as iterators to lists of elements to insert, and "hints" as to the position an element belongs. The use of a "hint" iterator can bring the time insertion takes down to contant time, otherwise it is O(log(N)) time.

Operator[]: Checks to see if the object exists, and if it does, modifies the reference to this object, otherwise uses the provided key and value to call make_pair on the two objects, and then does the same work as the insert function. This is O(log(N)) time.

make_pair: Does little more than make a pair.

There was no "need" for adding emplace to the standard. In c++11 I believe the && type of reference was added. This removed the necessity for move semantics, and allowed optimization of some specific type of memory management. In particular, the rvalue reference. The overloaded insert(value_type &&) operator does not take advantage of the in_place semantics, and is therefore much less efficient. While it provides the capability of dealing with rvalue references, it ignores their key purpose, which is in place construction of objects.

Answer 3

Apart from the optimisation opportunities and the simpler syntax, an important distinction between insertion and emplacement is that the latter allows explicit conversions. (This is across the entire standard library, not just for maps.)

Here's an example to demonstrate:

#include <vector>

struct foo
{
    explicit foo(int);
};

int main()
{
    std::vector<foo> v;

    v.emplace(v.end(), 10);      // Works
    //v.insert(v.end(), 10);     // Error, not explicit
    v.insert(v.end(), foo(10));  // Also works
}

This is admittedly a very specific detail, but when you're dealing with chains of user-defined conversions, it's worth keeping this in mind.

Answer 4

In terms of functionality or output, they are both same.

For both large memory, object emplace is memory-optimized which don't use copy constructors

For simple detailed explanation https://medium.com/@sandywits/all-about-emplace-in-c-71fd15e06e44

Answer 5

The following code may help you understand the "big picture idea" of how insert() differs from emplace():

#include <iostream>
#include <unordered_map>
#include <utility>

//Foo simply outputs what constructor is called with what value.
struct Foo {
  static int foo_counter; //Track how many Foo objects have been created.
  int val; //This Foo object was the val-th Foo object to be created.

  Foo() { val = foo_counter++;
    std::cout << "Foo() with val:                " << val << '\n';
  }
  Foo(int value) : val(value) { foo_counter++;
    std::cout << "Foo(int) with val:             " << val << '\n';
  }
  Foo(Foo& f2) { val = foo_counter++;
    std::cout << "Foo(Foo &) with val:           " << val
              << " \tcreated from:      \t" << f2.val << '\n';
  }
  Foo(const Foo& f2) { val = foo_counter++;
    std::cout << "Foo(const Foo &) with val:     " << val
              << " \tcreated from:      \t" << f2.val << '\n';
  }
  Foo(Foo&& f2) { val = foo_counter++;
    std::cout << "Foo(Foo&&) moving:             " << f2.val
              << " \tand changing it to:\t" << val << '\n';
  }
  ~Foo() { std::cout << "~Foo() destroying:             " << val << '\n'; }

  Foo& operator=(const Foo& rhs) {
    std::cout << "Foo& operator=(const Foo& rhs) with rhs.val: " << rhs.val
              << " \tcalled with lhs.val = \t" << val
              << " \tChanging lhs.val to: \t" << rhs.val << '\n';
    val = rhs.val;
    return *this;
  }

  bool operator==(const Foo &rhs) const { return val == rhs.val; }
  bool operator<(const Foo &rhs)  const { return val < rhs.val;  }
};

int Foo::foo_counter = 0;

//Create a hash function for Foo in order to use Foo with unordered_map
namespace std {
   template<> struct hash<Foo> {
       std::size_t operator()(const Foo &f) const {
           return std::hash<int>{}(f.val);
       }
   };
}

int main()
{
    std::unordered_map<Foo, int> umap;  
    Foo foo0, foo1, foo2, foo3;
    int d;

    //Print the statement to be executed and then execute it.

    std::cout << "\numap.insert(std::pair<Foo, int>(foo0, d))\n";
    umap.insert(std::pair<Foo, int>(foo0, d));
    //Side note: equiv. to: umap.insert(std::make_pair(foo0, d));

    std::cout << "\numap.insert(std::move(std::pair<Foo, int>(foo1, d)))\n";
    umap.insert(std::move(std::pair<Foo, int>(foo1, d)));
    //Side note: equiv. to: umap.insert(std::make_pair(foo1, d));

    std::cout << "\nstd::pair<Foo, int> pair(foo2, d)\n";
    std::pair<Foo, int> pair(foo2, d);

    std::cout << "\numap.insert(pair)\n";
    umap.insert(pair);

    std::cout << "\numap.emplace(foo3, d)\n";
    umap.emplace(foo3, d);

    std::cout << "\numap.emplace(11, d)\n";
    umap.emplace(11, d);

    std::cout << "\numap.insert({12, d})\n";
    umap.insert({12, d});

    std::cout.flush();
}

The output that I got was:

Foo() with val:                0
Foo() with val:                1
Foo() with val:                2
Foo() with val:                3

umap.insert(std::pair<Foo, int>(foo0, d))
Foo(Foo &) with val:           4    created from:       0
Foo(Foo&&) moving:             4    and changing it to: 5
~Foo() destroying:             4

umap.insert(std::move(std::pair<Foo, int>(foo1, d)))
Foo(Foo &) with val:           6    created from:       1
Foo(Foo&&) moving:             6    and changing it to: 7
~Foo() destroying:             6

std::pair<Foo, int> pair(foo2, d)
Foo(Foo &) with val:           8    created from:       2

umap.insert(pair)
Foo(const Foo &) with val:     9    created from:       8

umap.emplace(foo3, d)
Foo(Foo &) with val:           10   created from:       3

umap.emplace(11, d)
Foo(int) with val:             11

umap.insert({12, d})
Foo(int) with val:             12
Foo(const Foo &) with val:     13   created from:       12
~Foo() destroying:             12

~Foo() destroying:             8
~Foo() destroying:             3
~Foo() destroying:             2
~Foo() destroying:             1
~Foo() destroying:             0
~Foo() destroying:             13
~Foo() destroying:             11
~Foo() destroying:             5
~Foo() destroying:             10
~Foo() destroying:             7
~Foo() destroying:             9

Notice that:

1. An unordered_map always internally stores Foo objects (and not, say, Foo *s) as keys, which are all destroyed when the unordered_map is destroyed. Here, the unordered_map's internal keys were foos 13, 11, 5, 10, 7, and 9.

   • So technically, our unordered_map actually stores std::pair<const Foo, int> objects, which in turn store the Foo objects. But to understand the "big picture idea" of how emplace() differs from insert() (see highlighted box below), it's okay to temporarily imagine this std::pair object as being entirely passive. Once you understand this "big picture idea," it's important to then back up and understand how the use of this intermediary std::pair object by unordered_map introduces subtle, but important, technicalities.

2. Inserting each of foo0, foo1, and foo2 required 2 calls to one of Foo's copy/move constructors and 2 calls to Foo's destructor (as I now describe):

   a. Inserting each of foo0 and foo1 created a temporary object (foo4 and foo6, respectively) whose destructor was then immediately called after the insertion completed. In addition, the unordered_map's internal Foos (which are Foos 5 and 7) also had their destructors called when the unordered_map was destroyed.

   b. To insert foo2, we instead first explicitly created a non-temporary pair object (called pair), which called Foo's copy constructor on foo2 (creating foo8 as an internal member of pair). We then insert()ed this pair, which resulted in unordered_map calling the copy constructor again (on foo8) to create its own internal copy (foo9). As with foos 0 and 1, the end result was two destructor calls for this insertion with the only difference being that foo8's destructor was called only when we reached the end of main() rather than being called immediately after insert() finished.

3. Emplacing foo3 resulted in only 1 copy/move constructor call (creating foo10 internally in the unordered_map ) and only 1 call to Foo's destructor. (I'll get back to this later).

4. For foo11, we directly passed the integer 11 to emplace(11, d) so that unordered_map would call the Foo(int) constructor while execution is within its emplace() method. Unlike in (2) and (3), we didn't even need some pre-exiting foo object to do this. Importantly, notice that only 1 call to a Foo constructor occurred (which created foo11).

5. We then directly passed the integer 12 to insert({12, d}). Unlike with emplace(11, d) (which recall resulted in only 1 call to a Foo constructor), this call to insert({12, d}) resulted in two calls to Foo's constructor (creating foo12 and foo13).

This shows what the main "big picture" difference between insert() and emplace() is:

Whereas using insert() almost always requires the construction or existence of some Foo object in main()'s scope (followed by a copy or move), if using emplace() then any call to a Foo constructor is done entirely internally in the unordered_map (i.e. inside the scope of the emplace() method's definition). The argument(s) for the key that you pass to emplace() are directly forwarded to a Foo constructor call within unordered_map::emplace()'s definition (optional additional details: where this newly constructed object is immediately incorporated into one of unordered_map's member variables so that no destructor is called when execution leaves emplace() and no move or copy constructors are called).

Note: The reason for the "almost" in "almost always" above is explained in I) below.

3. continued: The reason why calling umap.emplace(foo3, d) called Foo's non-const copy constructor is the following: Since we're using emplace(), the compiler knows that foo3 (a non-const Foo object) is meant to be an argument to some Foo constructor. In this case, the most fitting Foo constructor is the non-const copy constructor Foo(Foo& f2). This is why umap.emplace(foo3, d) called a copy constructor while umap.emplace(11, d) did not.

Epilogue:

I. Note that one overload of insert() is actually equivalent to emplace(). As described in this cppreference.com page, the overload template<class P> std::pair<iterator, bool> insert(P&& value) (which is overload (2) of insert() on this cppreference.com page) is equivalent to emplace(std::forward<P>(value)).

II. Where to go from here?

a. Play around with the above source code and study documentation for insert() (e.g. here) and emplace() (e.g. here) that's found online. If you're using an IDE such as eclipse or NetBeans then you can easily get your IDE to tell you which overload of insert() or emplace() is being called (in eclipse, just keep your mouse's cursor steady over the function call for a second). Here's some more code to try out:

std::cout << "\numap.insert({{" << Foo::foo_counter << ", d}})\n";
umap.insert({{Foo::foo_counter, d}});
//but umap.emplace({{Foo::foo_counter, d}}); results in a compile error!

std::cout << "\numap.insert(std::pair<const Foo, int>({" << Foo::foo_counter << ", d}))\n";
umap.insert(std::pair<const Foo, int>({Foo::foo_counter, d}));
//The above uses Foo(int) and then Foo(const Foo &), as expected. but the
// below call uses Foo(int) and the move constructor Foo(Foo&&). 
//Do you see why?
std::cout << "\numap.insert(std::pair<Foo, int>({" << Foo::foo_counter << ", d}))\n";
umap.insert(std::pair<Foo, int>({Foo::foo_counter, d}));
//Not only that, but even more interesting is how the call below uses all 
// three of Foo(int) and the Foo(Foo&&) move and Foo(const Foo &) copy 
// constructors, despite the below call's only difference from the call above 
// being the additional { }.
std::cout << "\numap.insert({std::pair<Foo, int>({" << Foo::foo_counter << ", d})})\n";
umap.insert({std::pair<Foo, int>({Foo::foo_counter, d})});


//Pay close attention to the subtle difference in the effects of the next 
// two calls.
int cur_foo_counter = Foo::foo_counter;
std::cout << "\numap.insert({{cur_foo_counter, d}, {cur_foo_counter+1, d}}) where " 
  << "cur_foo_counter = " << cur_foo_counter << "\n";
umap.insert({{cur_foo_counter, d}, {cur_foo_counter+1, d}});

std::cout << "\numap.insert({{Foo::foo_counter, d}, {Foo::foo_counter+1, d}}) where "
  << "Foo::foo_counter = " << Foo::foo_counter << "\n";
umap.insert({{Foo::foo_counter, d}, {Foo::foo_counter+1, d}});


//umap.insert(std::initializer_list<std::pair<Foo, int>>({{Foo::foo_counter, d}}));
//The call below works fine, but the commented out line above gives a 
// compiler error. It's instructive to find out why. The two calls
// differ by a "const".
std::cout << "\numap.insert(std::initializer_list<std::pair<const Foo, int>>({{" << Foo::foo_counter << ", d}}))\n";
umap.insert(std::initializer_list<std::pair<const Foo, int>>({{Foo::foo_counter, d}}));

You'll soon see that which overload of the std::pair constructor (see reference) ends up being used by unordered_map can have an important effect on how many objects are copied, moved, created, and/or destroyed as well as when this all occurs.

b. See what happens when you use some other container class (e.g. std::set or std::unordered_multiset) instead of std::unordered_map.

c. Now use a Goo object (just a renamed copy of Foo) instead of an int as the range type in an unordered_map (i.e. use unordered_map<Foo, Goo> instead of unordered_map<Foo, int>) and see how many and which Goo constructors are called. (Spoiler: there is an effect but it isn't very dramatic.)