r175326 - Rework the visibility computation algorithm in preparation

[Rafael Espíndola]

> I found one more bug, with problem (1). It looks like the first
> declaration of a type must be marked visible, not just any declaration
> before an instantiation with that type:

The problem comes from the template using the canonical (first) decl
for the arguments. Some extra examples:

template <typename T>
struct __attribute__((visibility("default"))) barT {
  static void zed() {}
};
class foo;
class __attribute__((visibility("default"))) foo;
template struct barT<foo>;

template <int* I>
struct __attribute__((visibility("default"))) barI {
  static void zed() {}
};
extern int I;
extern int I __attribute__((visibility("default")));
template struct barI<&I>;

typedef void (*fType)(void);
template<fType F>
struct __attribute__((visibility("default"))) barF {
  static void zed() {}
};
void F();
void F() __attribute__((visibility("default")));;
template struct barF<F>;

This is somewhat easy to hack in the visibility computation (see
attached patch), but what should we do for cases like:


template <typename T>
struct __attribute__((visibility("default"))) barT {
  static void zed() {}
};
class foo;
template struct barT<foo>;
class __attribute__((visibility("default"))) foo;


With this patch we print a hidden visibility. But for

emplate <int* I>
struct __attribute__((visibility("default"))) barI {
  static void zed() {}
};
extern int I;
void use() {
  barI<&I>::zed();
}
extern int I __attribute__((visibility("default")));

We produce a default one unless the last line is commented. Should the
template instantiation point to the last decl available when it is
created?

Cheers,
Rafael
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