In S s = S() is it guaranteed that no temporary will be created?

Question

In the following code, are pS and s.pS guaranteed to be equal in the final line? In other words, in the statement S s = S(); , can I be sure that a temporary S will not be constructed?

#include <iostream>
using namespace std;

struct S
{
  S() { pS = this; }
  S* pS;
};

int main()
{
  S s = S();
  S* pS = &s;
  cout << pS << " " << s.pS << endl;
}

In every compiler I've tested this in pS == s.pS , but I'm not sufficiently familiar with the standard to be able to satisfy myself that this is guaranteed.

Answer 1

NO

The compiler isn't obligated to do copy elision. The standard simply specifies that, [class.copy]:

When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object [...]

I can disable copy elision via -fno-elide-constructors , and then the two pointers will definitely be different. For example:

$g++ -std=c++11 -Wall -pedantic -fno-elide-constructors -Wall -Wextra main.cpp && ./a.out
0x7fff5a598920 0x7fff5a598930

And in the general case, if we add S(S&& ) = delete , then the above code wouldn't even compile.

Answer 2

Most compilers performs what's called copy/move elision , which is specified by the C++ standard. But it is not guaranteed. For example, you can compile with -fno-elide-constructors in gcc and you'll see all constructors in all their glory.

Live example on Coliru

Answer 3

There is no guarantee that there will be no temporary. But the Big Three compilers will optimize it out (even with the -O0 switch).

To guarantee no temporary at all just write:

int main()
{
  // ...
  S s{};
  // ...
}

Or simply S s; .