RE: Question for ISO C standards gurus

["Dave Korn" <dk at artimi dot com>]

> -----Original Message-----
> From: gcc-owner Behalf Of Ian Lance Taylor
> Sent: 27 March 2004 01:26

> Diego Novillo writes:
> 
> > On Thu, 2004-03-25 at 13:08, Dave Korn wrote:
> > 
> > > "Here is the complete list of all sequence points in C++: 
[...snip...]
> > > after the evaluation of all function arguments in a function call 
> > > and before execution of any expressions in the function body

[...snip...]
> > So, the program below should print 5?  It certainly doesn't 
> with gcc 
> > 3.2 nor mainline:
> > 
> > int foo (int x)
> > {
> >   return x;
> > }
> > 
> > main ()
> > {
> >   int *p;
> >   int a[2];
> >   a[0] = 4;
> >   a[1] = 5;
> >   p = a;
> >   printf ("%d\n", foo (*p++));
> > }
> > 
> > It will print 5 if I use pre-increment on 'p', which is what I 
> > would've expected (I'm no language lawyer, though).
> 
> I can't see any reason why that would print 5.  Based on the 
> above, p should be incremented before the function call to 
> foo().  But that doesn't change the value of *p++, which 
> means "get the value at *p, then increment p".  The above 
> just means that this should not happen:
>     t1 = *p;
>     foo (t1);
>     p = p + 1;
> (at least, it should not be possible to observe that that 
> happened; in the above program, p, as a local variable whose 
> address is not taken, probably could be incremented after the 
> function call, since there is no way for anything to detect 
> when the increment occurred.)
> 
> Ian



Indeed, that's my understanding as well.  I'd also mention that if foo was a
nested function and had access from its inner scope to the value of *p in
the enclosing function, it should be 5 at the time control flow was in foo.



    cheers, 
      DaveK
-- 
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