# How to write a function to match RVO

See the Following code.
1. `get1`
* ```
Output:
100
----
0

    * We only call one constructor in this function.
2. `get2`
    * ```
      Output:
      100
      ----
      default
      move

* Call one constructor in A{100}, but call default + move constructor with `named variable`.

3. `get3`
* ```
Output:
default
----
default

    * Only call one constructor
#### Conclution
1. minimize your return branch. (Like `get3()`)
2. If you have multiple return branch, make the return object be constructor(unnamed) as many as posible. (Like `get1()`)
3. Although you need to do the thing like `get2`, it's ok because it's a move constructor.

#include <iostream>

struct A {
int m;
A() { std::cout << "default\n"; }
A(int i) : m{i} { std::cout << i << "\n"; }
A(const A&) { std::cout << "copy\n"; };
A(A&&) { std::cout << "move\n"; };
};

A get1(int argc) {
if (argc > -1) {
return A{100};
}
return A{0};
}
A get2(int argc) {
if (argc > -1) {
return A{100};
}
A a;
return a;
}
A get3(int argc) {
A a;
if (argc > -1) {
a.m=100;
}
a.m=0;
return a;
}

int main(int argc, char** argv) {
auto a = get1(argc);
std::cout << "----\n";
auto b = get1(argc - 100);
std::cout << "-------\n";
auto c = get2(argc);
std::cout << "----\n";
auto d = get2(argc - 100);
std::cout << "-------\n";
auto e= get3(argc);
std::cout << "----\n";
auto f = get3(argc - 100);
std::cout << "----\n";
return 0;
}