Prior to C++11 the following code created a temporary string object and passed it to push_back, which then copied it into the vector: vector<string> myVector;
MyVector.push_back("message") ;
As of C++11, member function push_back is now overloaded with a version that takes a(n) ________. This allows the preceding call to push_back to take the storage allocated for the temporary string and reuse it directly for the new element in the vector. The temporary string will be destroyed when the function returns, so there's no need for it to keep its content.

Question

Prior to C++11 the following code created a temporary string object and passed it to push_back, which then copied it into the vector: vector<string> myVector;
MyVector.push_back("message");
As of C++11, member function push_back is now overloaded with a version that takes a(n) ________. This allows the preceding call to push_back to take the storage allocated for the temporary string and reuse it directly for the new element in the vector. The temporary string will be destroyed when the function returns, so there's no need for it to keep its content.
A) const pointer
B) rvalue reference
C) rvalue pointer
D) const reference

Quizplus · Accepted Answer

As of C++11, push_back is overloaded with a version that takes an rvalue reference as its argument. This is the best choice in this scenario because an rvalue reference can bind to a temporary string object, allowing the storage allocated for the temporary string to be reused directly for the new element in the vector. This is more efficient than creating a new string object and copying the content of the temporary string into it.

Prior to C++11 the Following Code Created a Temporary String