I've been reading about (N)RVO and would like one, complete scenario description. I hope this question will serve other C++ -learners to clarify their ideas.
Suppose this scenario:
string get_string() {
string x("racecar");
//work on x...
return x;
}
string a( get_string() );
string b = get_string();
Please disregard the C++11 move-semantics for a moment.
- If no (N)RVO is performed, how many constructors/assignments/destructors will be
executed? (please point out, to which objects to they refer)
- What changes if (N)RVO is applied?
- Finally, how does the situation change in C++11 assuming, that
std::string
supports move-semantics.
1) Inside
get_string
, one string object (x) will be constructed using the constructor which takes aconst char*
.2) When the function returns, the string constructed inside will be copy constructed to a temporary string object in the caller's space.
3) The temporary will be copy constructed to
a
.4) See 1
5) See 2
6) See 3, but the copy will go to
b
With RVO, 2 an 5 can be eliminated by constructing the temporary inside the function via an invisible reference. With further copy elision (not RVO), 3 and 6 can be eliminated. So that leaves us with 2 constructions, both using the
const char*
constructor.With C++11 move semantics, the situation doesn't change at all if the compiler was good enough to get all the copy elision done. If copy elision isn't done, then 2, 3, 5 and 6 still exist, but become moves instead of copies. Unlike copy elision though, these moves are not an optional optimization. A conforming compiler must perform them, assuming it didn't already perform copy elision.