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From: "rafael.espindola at gmail dot com" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug c++/54399] New: Invalid partial change from dynamic to static initialization
Date: Wed, 29 Aug 2012 01:25:00 -0000
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=54399

             Bug #: 54399
           Summary: Invalid partial change from dynamic to static
                    initialization
    Classification: Unclassified
           Product: gcc
           Version: 4.8.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: rafael.espindola@gmail.com


I initially thought that this was a missed optimization in llvm
(http://llvm.org/pr13677), but Richard Smith convinced me this actually a bug
in gcc.

$ cat test1.cpp
struct foo {
  int a;
  int b;
  int c;
  int d;
  int e;
};
int zed();
int x = zed();
foo bar = {x, 1, 2, 3, 4};
$ cat test2.cpp
#include <stdio.h>
struct foo {
  int a;
  int b;
  int c;
  int d;
  int e;
};
extern foo bar;
int zed() { return bar.d; }
int main(void) {
  printf("%d\n", bar.a);
  return 0;
}

With gcc:

$ ~/gcc/build/gcc/xgcc -std=c++11 -B ~/gcc/build/gcc -c test1.cpp 
$ ~/gcc/build/gcc/xgcc -std=c++11 -B ~/gcc/build/gcc -c test2.cpp 
$ g++ test1.o test2.o -o t
$ ./t
3

and with clang:

$ clang test1.cpp test2.cpp -o t
[espindola@desktop llvm]$ ./t
0

If I understand Richard's argument correctly, the program must print 0 because

* from [basic.start.init] p2 both x and bar get dynamic initializations. In
that case x is initialized first and zed will see bar zero initialized.
* from p3, we can convert a variable to static initialization, but not part of
it. The possibilities are:
  1) Only x is converted to static initialization. x should be static
initialized to 0. When bar is dynamic initialized it will be initialized to {0,
1, 2, 3, 4}.
  2) Only bar is converted to static initialization. It should still be
initialized to {0, 1, 2, 3, 4}. In this case x is dynamically initialized to 3.
  3) Both x and bar are converted to static initialization. In this case x is 0
and bar is {0, 1, 2, 3, 4}.

The output produced by gcc doesn't match any of the possibilities provided by
the standard.