[cfe-dev] Query regarding global value initialization standard

[Karen Shaeffer]

On Wed, Sep 25, 2013 at 03:08:06PM -0700, Eli Friedman wrote:
> On Wed, Sep 25, 2013 at 3:02 PM, Karen Shaeffer <shaeffer at neuralscape.com>wrote:
> 
> > On Wed, Sep 25, 2013 at 02:40:37PM -0700, Eli Friedman wrote:
> > > On Wed, Sep 25, 2013 at 4:49 AM, Karthik Bhat <
> > blitz.opensource at gmail.com>wrote:
> > >
> > > > Hi All,
> > > > I was going through a gcc TC for C++11. The test case is as follows -
> > > >
> > > > // { dg-options -std=c++0x }
> > > > // { dg-do run }
> > > >
> > > > extern "C" void abort ();
> > > > extern int ar[2];
> > > >
> > > > int f()
> > > > {
> > > >   int k = 0;
> > > >   if (ar[0] != 42 || ar[1] != 0)
> > > >     abort();
> > > >   return 1;
> > > > }
> > > >
> > > > int i = f();
> > > >
> > > > int ar[2] = {42,i};
> > > >
> > > > int main()
> > > > {
> > > >   return 0;
> > > > }
> > > >
> > > > During dynamic initialization of i in function f() the value of ar[0]
> > is 0
> > > > in case of clang were as in case of gcc it is 42.
> > > >
> > > > As per standard(section 3.6.2) all global values should initially be
> > zero
> > > > initialized followed by const initialized if possible before dynamic
> > > > initialization takes place.
> > > > Hence as per standard the const initialization of int ar[2] = {42,i};
> > > > should fail as i is not a const here( which seems to be happening in
> > > > clang). Hence ar[0],ar[1] is still zero initialized because of which
> > during
> > > > dynamic initialization in f() ar[0]  is 0 which seems to be the correct
> > > > behavior.
> > > >
> > > > Can i conclude here that clang is behaving correctly and the tc is
> > wrong?
> > > > or am i missing something which this gcc tc wanted to capture?
> > > >
> > >
> > > As far as I can tell, your analysis is correct.
> > >
> > > -Eli
> >
> > Hi,
> > But isn't function f a constexpr? It doesn't modify anything and always
> > returns 1.
> >
> >
> It doesn't matter what the implementation of "f" is; "f()" is still not a
> constant expression.  Please read [basic.start.init] and [expr.const] in
> the C++ standard.
> 
> -Eli

Hi Eli,
OK. I agree. f() is not a constexpr because the body of the function is not a
return statement. There are other problems, but that is moot.

Based on my reading of the C++11 standard at 3.6.2 note 3, this g++ test case
appears to be fully compliant with the standard. I am referring to the static
initialization of ar[0] to 42. see notes below.

Temporary breakpoint 1, main (argc=1, argv=0x7fffffffe5c8) at constexpr.cpp:385
385     {
(gdb) print i
$1 = 1
(gdb) print ar
$2 = {42, 1}

When i is initialized by f(), ar[0] = 42 and ar[1] = 0. My interpretation is
that g++ is initializing ar[0] and ar[1] as individual variables, with respect to
the fact ar[1] has an initializer that is not a constant expression. Unless the
standard explicitly prohibits the static initialization of ar[0], because of some
all or nothing rule pertaining to the static initialization of the elements of an
array and noting i is not a constant expression, I believe the g++ test case is
valid code.

My interpretation doesn't necessarily mean clang has a bug here. But the standard
must explicitly forbid static initialization of ar[0], because the initializer of
ar[1] is not a constant expression, or at the least define it as implementation
dependent, or it would be a bug in clang. Where is this clarified in the standard?
Specifically, where in the standard does it require nonlocal static initialization
of elements of an array to be an all or nothing operation?


begin static initialization

i = 0
ar[0] = 0
ar[1] = 0
i ............. constant initialization is not done, because f() is not a constant
                expression

ar[0] = 42 .... constant initialization is done, because 42 is a constant literal
                integral type. Even if your interpreation is that ar[0] should be
                dynamically initialized, based on 3.6.2 note 3, ar[0] can be initialized
                either statically or dynamically. Its implementation dependent, unless
                the standard explicitly requires array element initialization to be
                an all or nothing operation.

ar[1] ......... constant initialization is not done, because i is not a constant
                expression

end static initialization
begin dynamic initialization
nonlocal variables with static storage duration have ordered dynamic initialization

i = f() = 1 ... f() does not abort because a[0] == 42 and a[1] == 0.
a[1] = 1

enjoy,
Karen
-- 
Karen Shaeffer                 Be aware: If you see an obstacle in your path,
Neuralscape Services           that obstacle is your path.        Zen proverb