From: JOAQUIN LOPEZ MU?Z (joaquin_at_[hidden])
Date: 2005-03-13 07:37:51
----- Mensaje original -----
De: David Abrahams <dave_at_[hidden]>
Fecha: Domingo, Marzo 13, 2005 4:52 am
Asunto: [boost] Re: [review] hash functions
> JOAQUIN LOPEZ MU?Z <joaquin_at_[hidden]> writes:
> > ----- Mensaje original -----
> > De: David Abrahams <dave_at_[hidden]>
> > Fecha: Domingo, Marzo 13, 2005 0:52 am
> > Asunto: [boost] Re: [review] hash functions
> >> Alberto Barbati <abarbati_at_[hidden]> writes:
> >> > Ah! And there's one more case in which we are helpless,
> >> I'm
> >> > curious to see an implementation that exploits such latitude:
> >> the
> >> > wording used in the standard does not guarantee that the
> >> from
> >> > pointers to integers always produce the same value over time.
> >> other
> >> > words:
> >> >
> >> > char* p = new char;
> >> > size_t i1 = reinterpret_cast<size_t>(p);
> >> > size_t i2 = reinterpret_cast<size_t>(p);
> >> > assert(i1 == i2); // ???
> >> > char* q1 = reinterpret_cast<char*>(i1);
> >> > char* q2 = reinterpret_cast<char*>(i2);
> >> > assert(p == q1 && q1 == q2);
> >> >
> >> > The standard guarantees that the second assert passes, but
> >> nothing
> >> > about the first assert.
> >> There are other cases too. reinterpret_cast could map all
> >> pointers to
> >> the same integer.
> > If you allow me to be a little pedantic, this is not the
> > case. reinterpret_cast<> guarantees back conversion, so it can't
> > possibly map every pointer into the same integer.
> 4 A pointer can be explicitly converted to any integral type large
> enough to hold it. The mapping function is implementation-defined
> [Note: it is intended to be unsurprising to those who know the
> addressing structure of the underlying machine. ]
> 5 A value of integral type or enumeration type can be explicitly
> converted to a pointer. 64) A pointer converted to an integer of
> sufficient size (if any such exists on the implementation) and back
> to the same pointer type will have its original value; mappings
> between pointers and integers are otherwise implementation-defined.
> There's no guarantee that the integer type will have sufficient size.
OK, with that provision in mind I buy your thesis now.
> > As for the rest of your argument, I agree 100%.
> Now you agree with all of it ;-)
> (I hope)
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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