--- In [email protected], "py2akv" <g...@...> wrote:
>
> Luciano,
>
> > const int i2 = 2;
> > int& ri2 = (int&) i2;
> > ri2 = 102; // should change the value of both 'i2' and 'ri2'
>
> Should it, my friend?
>
> As far as I'm concerned, ri2 is a reference type for an int, in this case
> const int i2 and it's correct the (int&) cast; if i2 weren't const, the cast
> wouldn't be necessary.
>
> Such being, i2 should never change, retaining the initialized value of 2.
>
> See there's nothing wrong with ri2 getting another value: it's not itself
> const; were ri2 const int& ri2 = (int&) i2, then the assignment ri2 = 102
> would be impossible.
>
> So, 102 affects ri2 only, but not i2, const.
But ri2 is a reference to i2. So whether i2=2 or i2=102, I would expect ri2=i2.
In fact if you modify the printfs to output the variable addresses as well:
printf("i2=%d %p\n", i2, &i2 ); // = 2 (???)
printf("ri2=%d %p\n", ri2, &ri2 ); // = 102 (OK)
I get (g++ under linux):
i2=2 0xbf95a1a8
ri2=102 0xbf95a1a8
This appears to show different values at the same address.
I suspect that because you are using a cast to overwrite a const value the
behaviour is undefined. However I'm curious to know what's actually happening -
optimisation perhaps?
However I'm only a C programmer, so apologies if I'm missing something obvious.
