Question

I came across this example from Accelerated C++

vector<string> func(const string&); //function declaration

vector<string> v;
string line = "abc";

v = func(line); //on entry, initialization of func's single parameter from line
         //on exit, both initialization of the return value and then assignment to v

My question is, since func takes a const string reference as a parameter, why is the copy constructor invoked when entering func? Since line is being passed by reference doesn’t func just keep a reference to line on its local stack?

Answer 1

on entry, initialization of func‘s single parameter from line

func‘s parameter is initialized from line, but it’s not a string, but a reference to it. Its initialization do not result in a call to the copy constructor, but it makes the parameter become an alias for line (as always happens with initialization of references).