Re: [Valgrind-users] "still reachable" memory from g++'sstd::string

[Olly B. <ol...@su...>]

On Wed, Apr 02, 2003 at 09:49:46AM -0800, Jeremy Fitzhardinge wrote:
> On Wed, 2003-04-02 at 08:34, Olly Betts wrote: 
> 
> > On Wed, Apr 02, 2003 at 11:23:34AM -0500, Maarten Ballintijn wrote:
> > > >         >           {
> > > >         >             std::string s;
> > > >         >             s += '0';
> > > >         >           }
> > > >         >           exit(0);
> > 
> > > Just out of curiosity, why would s be destructed if you call exit()?
> > 
> > s is destructed at the end of the block where it's defined - i.e. the
> > line before exit is called.
> 
> Do you mean the compiler is treating exit specially?

No.  Look at the code.  s is destructed at the end of the block it is
defined in.  That's how C++ works.  What happens after the block is
irrelevant to that destruction.

> Is that because
> its a noreturn function?  That still can't be right: if it were a
> noreturn function which takes a std::string argument, then it would be
> wrong to destruct the string before passing it.

But it couldn't take s as an argument, as s isn't in scope at that
point!

Cheers,
    Olly