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Is std::vector copying the objects with a push_back?

Tags:

c++

stl

stdvector

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What does the std::vector Push_back () method do?

The C++ function std::vector::push_back() inserts new element at the end of vector and increases size of vector by one.

Does std::vector copy?

The standard algorithm for copying is std::copy . We can use it for copying elements from the source vector to the destination vector.

Should I use Emplace_back or Push_back?

Specific use case for emplace_back : If you need to create a temporary object which will then be pushed into a container, use emplace_back instead of push_back . It will create the object in-place within the container. Notes: push_back in the above case will create a temporary object and move it into the container.


Yes, std::vector<T>::push_back() creates a copy of the argument and stores it in the vector. If you want to store pointers to objects in your vector, create a std::vector<whatever*> instead of std::vector<whatever>.

However, you need to make sure that the objects referenced by the pointers remain valid while the vector holds a reference to them (smart pointers utilizing the RAII idiom solve the problem).


From C++11 onwards, all the standard containers (std::vector, std::map, etc) support move semantics, meaning that you can now pass rvalues to standard containers and avoid a copy:

// Example object class.
class object
{
private:
    int             m_val1;
    std::string     m_val2;

public:
    // Constructor for object class.
    object(int val1, std::string &&val2) :
        m_val1(val1),
        m_val2(std::move(val2))
    {

    }
};

std::vector<object> myList;

// #1 Copy into the vector.
object foo1(1, "foo");
myList.push_back(foo1);

// #2 Move into the vector (no copy).
object foo2(1024, "bar");
myList.push_back(std::move(foo2));

// #3 Move temporary into vector (no copy).
myList.push_back(object(453, "baz"));

// #4 Create instance of object directly inside the vector (no copy, no move).
myList.emplace_back(453, "qux");

Alternatively you can use various smart pointers to get mostly the same effect:

std::unique_ptr example

std::vector<std::unique_ptr<object>> myPtrList;

// #5a unique_ptr can only ever be moved.
auto pFoo = std::make_unique<object>(1, "foo");
myPtrList.push_back(std::move(pFoo));

// #5b unique_ptr can only ever be moved.
myPtrList.push_back(std::make_unique<object>(1, "foo"));

std::shared_ptr example

std::vector<std::shared_ptr<object>> objectPtrList2;

// #6 shared_ptr can be used to retain a copy of the pointer and update both the vector
// value and the local copy simultaneously.
auto pFooShared = std::make_shared<object>(1, "foo");
objectPtrList2.push_back(pFooShared);
// Pointer to object stored in the vector, but pFooShared is still valid.

Yes, std::vector stores copies. How should vector know what the expected life-times of your objects are?

If you want to transfer or share ownership of the objects use pointers, possibly smart pointers like shared_ptr (found in Boost or TR1) to ease resource management.


std::vector always makes a copy of whatever is being stored in the vector.

If you are keeping a vector of pointers, then it will make a copy of the pointer, but not the instance being to which the pointer is pointing. If you are dealing with large objects, you can (and probably should) always use a vector of pointers. Often, using a vector of smart pointers of an appropriate type is good for safety purposes, since handling object lifetime and memory management can be tricky otherwise.


Not only does std::vector make a copy of whatever you're pushing back, but the definition of the collection states that it will do so, and that you may not use objects without the correct copy semantics within a vector. So, for example, you do not use auto_ptr in a vector.


Relevant in C++11 is the emplace family of member functions, which allow you to transfer ownership of objects by moving them into containers.

The idiom of usage would look like

std::vector<Object> objs;

Object l_value_obj { /* initialize */ };
// use object here...

objs.emplace_back(std::move(l_value_obj));

The move for the lvalue object is important as otherwise it would be forwarded as a reference or const reference and the move constructor would not be called.