Does passing an std::optional<T> by reference actually save copying?-CodePudding

I know that std::optional<T&> isn't supported in the standard. This question is about whether passing std::optional<T>& has any performance advantage

Sample code (https://godbolt.org/z/h56Pj6d6z) reproduced here

#include <ctime>
#include <iomanip>
#include <iostream>
#include <optional>

void DoStuff(std::optional<std::string> str) {
    if (str) std::cout << "cop: " << *str << std::endl;
}

void DoStuffRef(const std::optional<std::string>& str) {
    if (str) std::cout << "ref: " << *str << std::endl;
}

int main() {
    std::optional<std::string> str = {};
    DoStuff(str);
    DoStuffRef(str);
    str = "0123456789012345678901234567890123456789";
    DoStuff(str);
    DoStuffRef(str);
}

(My actual use case is an optional for a complex user-defined type, but I hope that a long string would do the same compiler-wise)

In this case, does DoStuffRef actually save any copying effort compared to DoStuff?

I tried to look at godbolt output but I don't know enough assembly to be sure. I do see that in the case of DoStuff, there seems to be a temp std::optional<T> created which is not present in DoStuffRef so my suspicion is that yes, passing an optional by reference save some copying

Appreciate the help!

CodePudding user response：

If you pass actual std::optional<std::string> then yes, there would be no copy. But if you pass just std::string then temporary optional has to be constructed first, resulting in a copy of the string.

CodePudding user response：

DoStuffRef saves an extra copy when argument is already std::optional<std::string> (as in your example).

But, if you pass directly a std::string, then if both cases, a std::optional should be constructed, involving copy/move constructor of the string.