Re: [sdcc-devel] pointer arithmatic unsigned?

[Bernhard H. <ber...@be...>]

> Signedness of "char" is implementation defined , hence 0xff can be
> treated as 256 & -1 depending on the implementation and both are correct
> no code should rely on this.
255 & -1. I fully agree with the rest. But what behaviour do you recommend
for:

signed char sc;
unsigned char uc;
char *cp;

void foo (void) {
  cp += sc;    // sign/zero extension?
  cp += uc;    // sign/zero extension?
}

Bernhard

>
> Sandeep
>
> > -----Original Message-----
> > From: Bernhard Held [mailto:ber...@be...]
> > Sent: Wednesday, February 13, 2002 11:08 PM
> > To: Sandeep Dutta; 'Johan Knol'; sdc...@li...
> > Subject: Re: [sdcc-devel] pointer arithmatic unsigned?
> >
> >
> > > No .. pointer arithmetic is always unsigned ..
> > > sandeep
> > No, sorry, no, this can't be true. It must be possible to decrement a
> > pointer with a signed char or short. I've tested several
> > compilers. All of
> > them perform a sign extension with signed integers.
> >
> > > > char c;
> > > > char *cp;
> > > >
> > > > void foo (void) {
> > > >     cp += c;
> > > > }
> > > >
> > > >
> > > > Should c be sign-extended before being added to cp?
> >
> > signed char c;    sign extension
> > unsigned char c;    zero extension
> >
> > Bernhard
> >
> >
> >
>
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