Context: I was making experiments to learn when does gcc perform RVO, and if not, when does it use move semantics. My version of gcc is g++ (GCC) 4.8.5 20150623 (Red Hat 4.8.5-4)
.
Question: I have a function that returns a Foo
by value. Compiler cannot perform RVO because there are two possible named return values. When I use the ternary operator to select which of Foo
to return, then I need to explicitly call std::move
to avoid the copy. I do not need the std::move
when using an if statement, I do not need to. Why the discrepancy?
Code:
#include <iostream>
using namespace std;
struct Foo {
std::string s;
Foo() { cout << "Foo()\n"; }
~Foo() { cout << "~Foo()\n"; }
Foo(const Foo& other) : s(other.s) { cout << "Foo(const Foo&)\n"; }
Foo(Foo&& other) noexcept : s(move(other.s)) { cout << "Foo(Foo&&)\n"; }
};
Foo makeFooIf(bool which) {
Foo foo1; foo1.s = "Hello, World1!";
Foo foo2; foo2.s = "Hello, World2!";
if (which) return foo1;
else return foo2;
}
Foo makeFooTernary(bool which) {
Foo foo1; foo1.s = "Hello, World1!";
Foo foo2; foo2.s = "Hello, World2!";
return which ? foo1 : foo2;
}
Foo makeFooTernaryMove(bool which) {
Foo foo1; foo1.s = "Hello, World1!";
Foo foo2; foo2.s = "Hello, World2!";
return which ? move(foo1) : move(foo2);
}
int main()
{
cout << "----- makeFooIf -----\n";
Foo fooIf = makeFooIf(true);
cout << fooIf.s << endl;
cout << "\n----- makeFooTernary -----\n";
Foo fooTernary = makeFooTernary(true);
cout << fooTernary.s << endl;
cout << "\n----- makeFooTernaryMove -----\n";
Foo fooTernaryMove = makeFooTernaryMove(true);
cout << fooTernaryMove.s << endl;
cout << "\n----- Cleanup -----\n";
return 0;
}
Output:
----- makeFooIf -----
Foo()
Foo()
Foo(Foo&&)
~Foo()
~Foo()
Hello, World1!
----- makeFooTernary -----
Foo()
Foo()
Foo(const Foo&)
~Foo()
~Foo()
Hello, World1!
----- makeFooTernaryMove -----
Foo()
Foo()
Foo(Foo&&)
~Foo()
~Foo()
Hello, World1!
----- Cleanup -----
~Foo()
~Foo()
~Foo()
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