curry@hplred.HP.COM (Bo Curry) (11/16/90)
A simple question for the wizards:
Assume I define
union {
FOOTYPE a[4];
BARTYPE b;
} combo;
Does the standard guarantee that (void *)(&combo.a[0]) == (void *)(&combo.b) ?
Thanks,
Bo Curry
curry@hplabs.hpl.hp.comgwyn@smoke.brl.mil (Doug Gwyn) (11/17/90)
In article <810001@hplred.HP.COM> curry@hplred.HP.COM (Bo Curry) writes: >Does the standard guarantee that (void *)(&combo.a[0]) == (void *)(&combo.b) ? Yes, as a consequence of a requirement in 3.5.2.1 Semantics combined with the property of void* as a universal object pointer type.
henry@zoo.toronto.edu (Henry Spencer) (11/18/90)
In article <810001@hplred.HP.COM> curry@hplred.HP.COM (Bo Curry) writes: >union { > FOOTYPE a[4]; > BARTYPE b; >} combo; > >Does the standard guarantee that (void *)(&combo.a[0]) == (void *)(&combo.b) ? It guarantees (3.5.2.1) that a pointer to the union points to each of its members, after suitable conversion. That is not *quite* what you are asking, but I think it would take a seriously bizarre implementation to do otherwise. It may be possible to prove your equality, although it would take non-trivial reasoning. -- "I don't *want* to be normal!" | Henry Spencer at U of Toronto Zoology "Not to worry." | henry@zoo.toronto.edu utzoo!henry
steve@taumet.com (Stephen Clamage) (11/18/90)
curry@hplred.HP.COM (Bo Curry) writes:
!union {
! FOOTYPE a[4];
! BARTYPE b;
!} combo;
!Does the standard guarantee that (void *)(&combo.a[0]) == (void *)(&combo.b) ?
Not exactly. The standard (section 3.5.2.1) requires that
(FOOTYPE*)&combo == &combo.a[0]
and that
(BARTYPE*)&combo == &combo.b
There is also a requirement that given a type T, and the declaration
T* t;
then
(T*)(void*)t == t;
I don't believe that there is any requirement that the relation you specify
must hold. On machines where pointers to different types may have
different sizes, their conversions to void* might not be equal. The
standard only requires that conversion from void* back to the original
pointer type must work.
--
Steve Clamage, TauMetric Corp, steve@taumet.comsteve@taumet.com (Stephen Clamage) (11/18/90)
steve@taumet.com (Stephen Clamage) writes: >There is also a requirement that given a type T, and the declaration > T* t; >then > (T*)(void*)t == t; I didn't state this very well. I meant to say: There is also a requirement that given an object type T, and the declaration T* t; then the relation (T*)(void*)t == t must hold. -- Steve Clamage, TauMetric Corp, steve@taumet.com
msb@sq.sq.com (Mark Brader) (11/22/90)
> > union { > > FOOTYPE a[4]; > > BARTYPE b; > > } combo; > > Does the standard guarantee > > (void *)(&combo.a[0]) == (void *)(&combo.b) ? > > Not exactly. The standard (section 3.5.2.1) requires that > (FOOTYPE*)&combo == &combo.a[0] No, combo.a is the union member, not combo.a[0]. So this should be: (FOOTYPE(*)[4])&combo == &combo.a The actual wording in 3.5.2.1 is: # A pointer to a union object, suitably converted, points to each # of its members (or if the member is a bit-field, then to the unit # in which it resides), and vice versa. One sticking point is the meaning of "suitably converted". When I casually read this before, I assumed that converting either pointer to void * would be "suitable". (The other pointer in the comparison would then be implicitly converted to void * also.) In that case, &combo == (void *)&combo.a and &combo == (void *)&combo.b and therefore (void *)&combo.a == (void *)&combo.b But the stricter interpretation, guaranteeing only (FOOTYPE(*)[4])&combo == &combo.a and (BARTYPE *)&combo == &combo.b and &combo = (union ... *)&combo.a and &combo = (union ... *)&combo.b also seems to be legitimate. We'd have to have an interpretation ruling to settle this, I think. Anyway, note the last two equalities. They imply that (void *)(union ... *)&combo.a == (void *)(union ... *)&combo.b but, as there isn't any direct guarantee that (void *)(TYPE *)foop == (void *)foop , this probably doesn't help. Likewise, there is no guarantee that a pointer to an array, suitably converted, points to its first element, or vice versa. That is, even if (void *)&combo.a == (void *)&combo.b is true, the equality originally asked about does not seem to follow. It is, of course, possible that some of these things are indirectly guaranteed, and I haven't noticed. -- Mark Brader "'Settlor', (i) in relation to a testamentary trust, Toronto means the individual referred to in paragraph (i)." utzoo!sq!msb, msb@sq.com -- Income Tax Act of Canada, 108(1)(h) This article is in the public domain.