Compiler error in VS2013 and VS2015?

Question

In the following code the expression marked evaluates differently if compiled with VS6SP6 and with VS2013SP4/VS2015preview.
I suspect the code generated by VS2013/VS2015 is wrong. It creates a temporary object of *pa instead of using *pa (as it is done by VS6).
Or is it correct in C++, that for the given expression the Compiler creates a temporary object?
#include "stdio.h"
class A
public:
A() { printf("constructor A: %4X\n", this); }
A(A& a) { printf("constructor A(copy): %4X => %4X\n", &a, this); }
~A() { printf("destructor A: %4X\n", this); }
int aA;
A* pa;
A* pb;
void pp(A& a)
if (&a == pa)
printf("Ok, got expected object\n");
else
printf("not Ok, got %4X, expected %4X\n", &a, pa);
int main(int argc, char* argv[])
int i = 1;
pa = new A();
pb = new A();
pp((i == 0) ? *pb : *pa);
pp((i == 0) ? A() : *pa); //evaluates differently in VS6 and VS2015
delete pb;
delete pa;
return 0;
output, if compiled with VS6SP6:
constructor A: 3A2F80
constructor A: 3A06E0
Ok, got expected object
Ok, got expected object
destructor A: 3A06E0
destructor A: 3A2F80
output, if compiled with VS2015 preview
constructor A: 13B8FC8
constructor A: 13B4398
Ok, got expected object
constructor A(copy): 13B8FC8 => 112F8E4
not Ok, got 112F8E4, expected 13B8FC8
destructor A: 112F8E4
destructor A: 13B4398
destructor A: 13B8FC8

Answer 1

On 1/20/2015 8:18 AM, MKrzikalla wrote:
  pp((i == 0) ? A() : *pa); //evaluates differently in VS6 and VS2015
VS6 is wrong, even by the text of the C++98 standard (the version of the standard contemporary with it):
5.16/5 Otherwise [when the second and third operand are not both lvalues - IT], the result is an rvalue...
5.16/6 Lvalue-to-rvalue (4.1) ... standard conversions are performed on the second and third operands...
12.2/1 Temporaries of class type are created in various contexts: ... a conversion that creates an rvalue (4.1, 5.2.9, 5.2.11, 5.4) ...
 The C++11 standard is much more explicit:
5.16/6 ...The second and third operands have the same type; the result is of that type. If the operands have class type, the result is a prvalue temporary of the result type, which is copy-initialized from either the second operand or the third operand depending
on the value of the first operand.
For further discussion, see
http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#446
(this is how the wording in C++11 came about).
Igor Tandetnik