On Thu, Feb 17, 2011 at 11:36:53AM +0000, Jonathan Wakely wrote:
> On 17 February 2011 09:44, Axel Freyn <axel-freyn@xxxxxx> wrote:
> > On Wed, Feb 16, 2011 at 01:37:03PM -0800, Bob Plantz wrote:
> >> On 02/16/2011 12:41 PM, Jason Mancini wrote:
> >>> Though I still find the output of this odd:
> >>>
> >>> for (char i(1); i>0; ++i)
> >>> printf("%d %d\n", i, sizeof(i));
> >>>
> >>> ...
> >>> 362195 1
> >>> 362196 1
> >>> 362197 1
> >>> ...
> >>>
> >>> For very large values of char! ^_^
> >>>
> >>> Jason
> >> That's odd. With g++ 4.4.5 on an x86-64 machine in 64-bit mode I get:
> >>
> >> ---
> >> 125 1
> >> 126 1
> >> 127 1
> > But there is also a second "bug" in the program. You tell "printf", that
> > the variable "i" is a integer and not a char -- so printf will read
> > sizeof(int) Bytes at the adress of "i" in order to create the
> > output-number, which gives you the 362195. You should write something
> > like
> > printf("%d %d\n", (int)i, sizeof(i));
> > in order to get the "true" value of i in the output -- then I would
> > expect values in the "char"-range -127<i<128.
>
> I'm guessing you didn't bother to actually compile that and test it?
Well, I did. But using gcc instead of g++. I get values <128 as soon as
I add the integer conversion (in C code).
> See 5.2.2 [expr.call] paragraph 7 in the C++ standard. Arguments to
> varargs functions are subject to integral promotions.
In C++ it does not make any difference.
Axel
