Why does an object returned by value have the same address as the object inside the method?

Question

#include <stdio.h>
#include <array>
#include <vector>

std::vector<int> foo() {
 int i;
 std::vector<int> a(100);
 printf("%p, %p, %p\n", &i, &a, &(a[0]));
 return a;
}

int main() {
 int i;
 std::vector<int> b = foo();
 printf("%p, %p, %p\n", &i, &b, &(b[0]));
}

Why do a and b have the same address for the above? Is this some kind of "cross-stack-frame" optimization? The result is the same even when I use the -O0 option.

The output:

$ vim main.cpp 
$ cc -std=c++11 -lc++ main.cpp
$ ./a.out
0x7ffee28d28ac, 0x7ffee28d28f0, 0x7ff401402c00
0x7ffee28d290c, 0x7ffee28d28f0, 0x7ff401402c00
$ 

Answer 1

This is because of copy elision/named return value optimization (NRVO). foo returns a named object a. So the compiler is not creating a local object and returning a copy of it but creates the object at the place where the caller puts it. You can read more about it on https://en.cppreference.com/w/cpp/language/copy_elision. While RVO is mandatory since C++17, NRVO is not, but it looks like your compiler supports it even in case of -O0.

Answer 2

Note that even without copy elision (mandatory or not), it's already possible for the addresses of the two objects to be the same because their lifetimes are non-overlapping.

Answer 3

as the others have explained this is because of copy elison. You can disable that with g++ by:

g++ -fno-elide-constructors main.cpp

where main.cpp contains your code.

now a and b have different addresses but a[0] and b[0] will have the same address because of the move semantics.