From: Rani Sharoni (rani_sharoni_at_[hidden])
Date: 2003-01-30 08:11:05
"Daniel Frey" <daniel.frey_at_[hidden]> wrote in message
news:3E38E30B.F8B43FCB_at_aixigo.de...
> > > Compiler: GNU C++ version 3.2 20020927 (prerelease)
> I also tried the GCC 3.2.1, but without success. It compiles, but it
> gives the wrong results.
> > > Any ideas, or results from other compilers?
>AFAICS the Intel 7 works fine.
> > I was able to complie the attached code (with minor improvment) using
strict
> > Comeau and VC7.1 and boost 1.29.0.
> One point about the implementation: The getC()-function isn't necessary,
Silly me. It seems that I got used to that function.
The following complied with VC6, VC7, VC7.1 and Comeau:
template <typename B, typename D>
struct is_base_and_derived
{
private:
typedef char (&yes)[1];
typedef char (&no) [2];
template <typename T>
static yes check(D const volatile *, T);
static no check(B const volatile *, int);
struct Host
{
operator B const volatile *() const;
operator D const volatile *();
};
public:
enum { result = sizeof(check(Host(), 0)) == sizeof(yes) };
};
struct B {};
struct B1 : B {};
struct B2 : B {};
struct D : B1, private B2 {};
typedef char Test[is_base_and_derived<B, D>::result];
typedef char Test[is_base_and_derived<B1,D>::result];
typedef char Test[is_base_and_derived<B2,D>::result];
typedef char Test[!is_base_and_derived<int,D>::result];
// Caveat VC6/7 BUG -
// typedef char Test[!is_base_and_derived<D,D>::result];
I was able to complie the following code using GCC 3.1 which proves that GCC
knows the required rules:
template<typename T>
long *f(D*, T);
char *f(B*, int);
struct Host1 {
operator B*() const;
operator D*();
};
long *x1 = f(Host1(), 0);
template<typename T>
long *g(B2*, T);
char *g(B1*, int);
struct Host2 {
operator B1*() const;
operator B2*();
};
char *x2 = g(Host2(), 0);
BCC 5.6 selected the wrong function (x1) which proves that it disagree about
the basic rules.
Rani