Question

Does a copy still take place here:

std::vector<int> &f = foo();

where foo‘s prototype is

std::vector<int> foo();

Answer 1

foo() returns a temporary object which cannot be bound to non-const reference.

Do one of these:

//temporary can be bound to const-reference, so this is ok
const std::vector<int> &f = foo();  //no copy takes place.

//or save a copy of temporary
std::vector<int> f = foo(); //copy takes place (may be optimized by compiler)

Read the comments above. In the second version (non-reference version), the compiler may optimize the return value, avoiding the copying of temporary. It depends on the implementation of foo() as well.

See this:

• Return value optimization