Re: C++ Conundrum?

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Tue, 13 Nov 2007 12:35:41 -0500
Message-ID:
<fhcn9f$h63$1@news.datemas.de>
Duncan Smith wrote:

I'm pondering one of those job interview style conundrums you see from
time to time, I think I have the answer, but it would be nice to back
it up against a specifcaton or something concrete to be sure...

Here goes:
<code snippet>
/// File A
#include <iostream>

extern int f(void);
extern int b;

int a = f();

int main(int argc, char* argv[])
{
   std::cout << b;
   return 0;
}

/// File B

extern int a;
int b=a;

int f()
{
   return 3;
}
</code snippet>

The problem description is that some compilers will correctly output
'3' whereas others may output garbage.


They may not output garbage.

 I guess this is because of the
depencies in the external linkage, b=a, a=f.

Perhaps the standard leaves the initialization order upto the
compiler/ linker implementation, so if Compiler A does a=f first
we'll get '3', but if another compiler sets b=a first, a will be
undefined so the output from cout will be the un-initialized memory
address...

Sound about right?


No compiler should produce garbage. It's either 3 or 0. Both 'a' and
'b' have static storage, which should be zero-initialised before any
dynamic initialisation (the order of which is unspecified) takes place.

So, you have two possibilities, scenario A and scenario B.

Scenario A: objects in TU 'A' are initialised first.
    -- static initialisation:
    a := 0
    b := 0
    --- dynamic initialisation
    a := 3 // by calling 'f'
    b := 3 // by assigning from external 'a'
    ---- main is called
    Output: 3

Scenario B: objects in TU 'B' are initialised first.
    -- static initialisation:
    a := 0
    b := 0
    --- dynamic initialisation
    b := 0 // by assigning from external 'a' (which is 0 now)
    a := 3 // by calling 'f'
    ---- main is called
    Output: 0

V
--
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