On Sun,  3 Sep 2000 12:30:15 GMT, assert@my-deja.com wrote in
comp.std.c++:

> I want to explore what the lower bound for bits is
> for an underlying type to represent an enumeration.
>=20
> enum E { a =3D 8, b =3D 15 };
>=20
> Can an integral type with 3 bits of representation
> be the underlying type if E ?

No.

> Or is it at least 4 ?

Yes.

> 7.2 p5 suggests that the underlying type will at
> least have enough bit patterns as there are enum values.

This can be confusing if taken in isolation.

> 7.2 p6
>=20
>   For an enumeration where e_min is the smallest enumerator and e_max i=
s
>   the largest, the values of the enumeration are the values of the
>   underlying type in the range b_min to b_max , where b_min and
>   b_max are, respectively, the smallest and largest values of
>   the smallest bit=ADfield that can store e_min and e_max.

Did you note footnote 81, on the same page as the paragraph above,
which describes what the terms b_min and b_max mean?

=3D=3D=3D=3D=3D=3D=3D=3D
81) On a two?s=ADcomplement machine, bmax is the smallest value greater
than or equal to max (abs(emin ) - 1 , abs(emax ) ) of the form 2^M
- 1; bmin is zero if emin is non=ADnegative and - (bmax + 1 ) otherwise.

> That last part is hard for me to get, what exactly is
>=20
> "the smallest bit field that can store e_min and e_max" ?

The question is not relevant.  The proper question is:

"what is the smallest bit field that can store b_min and b_max"?

If the minimum value of the enum is greater than 0, the bit field must
be large enough to store all values from 0 through e_emax, inclusive.

> does that mean if e_min is 8 and e_max is 15, that a
> 3 bit bit-field can store 8 to 15 since there is a mapping
> of the values 8 to 15 into the 3 bit bit-field ?
>=20
> 8  -> 000
> 9  -> 001
> ...
> 15 -> 111

No, this is not legal.  The bit field must be able to contain the
value 0 even if 0 is not one of the enumeration constants.  bmin is
never greater than 0.

> Or does that mean that all values from zero up
> to the value of e_max can must represented in the bitfield ?

Yes, and possibly more if any of the enumerated values are negative.

> If this is what it means then why not just say
>=20
> "the smallest bit-field that can store values from zero to e_max"
>=20
> Why even mention e_min ?

I believe Yaroslav already explained that adequately.  Consider an
enum with all negative values, for example -6 to -1.

> I tend to think that the first interpretation is correct
> (and that E may be represented with a 3 bit integral type)
> but I have seen posts regarding enum types that suggest
> the second interpretation, that is why I ask for clarification.
>=20
> As a final note I know that 3 bit integral types are hard
> to come by :) but I wanted my example to focus on the rules
> regarding enum and not distract from the issue by choosing a lot
> of non-obvious enum values that might express the
> same problem where the bit boundry is a 33 or 34 bit
> integral type.
>=20
> Thanks.

The issue about bit field sizes smaller than CHAR_BIT is not
far-fetched at all.  If you are defining a bit-field type containing
the enumeration value, it should indeed occupy no more than the
minimum number of bits specified by the standard.

Jack Klein
--=20
Home: http://jackklein.home.att.net

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I want to explore what the lower bound for bits is
for an underlying type to represent an enumeration.

enum E { a = 8, b = 15 };

Can an integral type with 3 bits of representation
be the underlying type if E ?

Or is it at least 4 ?

7.2 p5 suggests that the underlying type will at
least have enough bit patterns as there are enum values.

7.2 p6

  For an enumeration where e_min is the smallest enumerator and e_max is
  the largest, the values of the enumeration are the values of the
  underlying type in the range b_min to b_max , where b_min and
  b_max are, respectively, the smallest and largest values of
  the smallest bit   field that can store e_min and e_max.


That last part is hard for me to get, what exactly is

"the smallest bit field that can store e_min and e_max" ?

does that mean if e_min is 8 and e_max is 15, that a
3 bit bit-field can store 8 to 15 since there is a mapping
of the values 8 to 15 into the 3 bit bit-field ?

8  -> 000
9  -> 001
...
15 -> 111

Or does that mean that all values from zero up
to the value of e_max can must represented in the bitfield ?
If this is what it means then why not just say

"the smallest bit-field that can store values from zero to e_max"

Why even mention e_min ?

I tend to think that the first interpretation is correct
(and that E may be represented with a 3 bit integral type)
but I have seen posts regarding enum types that suggest
the second interpretation, that is why I ask for clarification.

As a final note I know that 3 bit integral types are hard
to come by :) but I wanted my example to focus on the rules
regarding enum and not distract from the issue by choosing a lot
of non-obvious enum values that might express the
same problem where the bit boundry is a 33 or 34 bit
integral type.

Thanks.


Sent via Deja.com http://www.deja.com/
Before you buy.

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