Re: [Devel] Off topic? Signed versus unsigned for hexadecimal constants.

[Antoine Leca]

Tom Kacvinsky wrote:
> 
> Based on the email earlier today about hexadecimal constants,

Note that I have *not* read the entire discussion with full attention...

> I have found the following (no qualifiers such as U or UL or L on the 
> constant):
<snip> 
> I don't have my ANSI C99 spec. in front of me (its at work, I'm at home), but 
> based on
> the above, I'd have to say its the ANSI specification that hexadecimal 
> constants should
> be treated as signed unless the U qualifier is given.

Nope. The rule is quite a bit more complex. Here they are:
For *decimal* integer constants:
 0 <= x <= INT_MAX                signed int
 INT_MAX < x <= LONG_MAX          signed long (if it matters)
 LONG_MAX < x <= ULONG_MAX        unsigned long
(and obviously, with C99 and the added types like long long, things are still
more complex, but detailing them is not worth the trouble)

For *octal* and *hexadecimal* integer constants:
 0 <= x <= INT_MAX                signed int
 INT_MAX < x <= UINT_MAX          unsigned int
 UINT_MAX < x <= LONG_MAX         signed long (if it matters)
 LONG_MAX < x <= ULONG_MAX        unsigned long (if it matters)

You can easily notice that with decimal constants, you cannot create a
"unsigned int"-typed value, so the "u" suffix is required. OTOH, with
octal and hexadecimal constants, you always have the "tallest" available
type with the correct value.

Now (assuming 32-bit ints) if 0xFFFFFFFF is signed (being int or long,
BTW), I would call it a bug.
(in ANSI mode, of course; things are quite different when one use non
conforming modes, for example because previous versions did not have these
rules, and preserving backward compatibility were and is considered more
important).


Antoine