: Re: [RFA] Setting long long bitfields

: Re: [RFA] Setting long long bitfields

[Paul Hilfinger]

Oops, sorry: I got the problematic case backwards.  The problem is
that a bitsize of 0 is legal, and that causes the result of your 
suggested tweak to be officially undefined.

Paul

>  > > One possible tweak, several of us have an aversion to "?:", it would be
>  > > nice if it wasn't there :-)
>  > 
>  > Like this perhaps:
>  > 
>  > +  ULONGEST mask = (ULONGEST) -1 >> (8 * sizeof (ULONGEST) - bitsize);
> 
> Andreas,
> 
> I had considered exactly that line, but unfortunately ran across the
> following really irritating provision in the C standard:
> 
>     "If the value of the right operand is negative or is 
>      *greater than or equal to* the width of the promoted left
>      operand, the behavior is undefined."
> 
> [I know why the provision is there, of course: many machines treat a
> shift of wordsize bits as 0, because they mask off bits to the left,
> but it's STILL irritating.]
> 
> Now, I presume we will never encounter a bitfield size of 8 * sizeof
> (ULONGEST), but the original code apparently tried to bullet-proof
> against this possibility, so I just went along.
> 
> Paul

Re: : Re: [RFA] Setting long long bitfields

[Andreas Schwab]

Paul Hilfinger <hilfingr@EECS.Berkeley.EDU> writes:

> Oops, sorry: I got the problematic case backwards.  The problem is
> that a bitsize of 0 is legal, and that causes the result of your 
> suggested tweak to be officially undefined.

How can bitsize ever become zero?  Note that the original code contains a
shift by bitsize - 1, which is undefined as well in this case.

Andreas.

-- 
Andreas Schwab, SuSE Labs, schwab@suse.de
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