Re: Enumeration representation clause surprise.

[Robert A Duff <bobduff@shell01.TheWorld.com>]

Adam Beneschan <adam@irvine.com> writes:

>...So
> what number is stored in a 0-bit data item?

Zero.

An n-bit field is either signed or unsigned.
When you load the field into a register, you
take all the bits and add copies of the sign bit on the left
(for signed) or zeros on the left (for unsigned).

A 0-bit field must be unsigned (there's no sign bit!).
So you take nothing, and add 32 zero bits on the left
(on a 32-bit machine).

Or you could think of it as an infinitely-long string of bits -- you
sign extend or zero extend infinitely on the left to get the value.

Then of course you can bias that 0 by anything -- range 100..100
can be stored in a zero-bit unsigned bitfield, biased by 100.
(Or is that -100?)

>...I suppose you could
> arbitrarily decide it's zero (which would make some sense
> mathematically)

Yes, I think so.

>... but it seems more commonsensical to say that nothing
> is stored there because there ain't no bits to store anything in.

But that view requires a special case.

The natural unsigned range for n bits, where n >= 0,
is 0..(2**n)-1.  Which is 0..0 when n = 0.

The natural signed (2's complement) range for n bits, where n > 0,
is -(2**(n-1))..(2**(n-1))-1.  Which makes no sense
if n = 0, because we don't want to mess around with
negative exponents (fractions) -- we're talking about
integer ranges.

- Bob