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/* APPLE LOCAL file C++ */
/* It should be possible to be possible to explicitly call 'operator ...'
functions with an implicit 'this', so long as the base class is not
a dependent type. For dependent base types, 'this->' or 'Base::'
must be used. */
/* Contributed by Ziemowit Laski <zlaski@apple.com> */
/* { dg-do compile } */
class Foo
{
public:
void read(int i);
};
void Func(Foo* inFoo);
void Foo::read(int i)
{
i = i + 1;
}
class Baz
{
public:
Baz(Foo* in) : fFoo(in) {}
operator Foo*() { return fFoo; }
Foo *foofoo(void) { return fFoo; }
Foo* fFoo;
};
template <class T> class Boop {
public:
Foo *booboo(void) { return gFoo; }
operator void *(void) { return (void *)gFoo; }
Foo *gFoo;
};
void Func(Foo* inFoo)
{
inFoo->read(1);
}
template <class T> class Bar : public Baz, public Boop <T>
{
public:
Bar(T *inFoo) : Baz(inFoo) {}
void CallMe()
{
void *p;
Func(this->foofoo());
Func(this->operator Foo*());
Func(this->booboo());
p = Boop<T>::operator void*();
Func(foofoo());
Func(operator Foo*());
Func(booboo()); /* { dg-error "must be available" } */
p = operator void *(); /* { dg-error "must be available" } */
}
};
int main (int argc, char * const argv[])
{
Foo theFoo;
Bar<Foo> theBar(&theFoo);
theBar.CallMe();
}
/* { dg-error "allowing the use of an undeclared name is deprecated" "" { target *-*-* } 0 } */
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