When we're checking a friend function template in an out-of-line class

definition, we may not have a scope corresponding to the namespace
where that friend function template actually lives. Work around this
issue by faking up a scope with the appropriate DeclContext.

This is a bit of a hack, but it fixes <rdar://problem/10204947>.

llvm-svn: 143614
This commit is contained in:
Douglas Gregor 2011-11-03 16:37:14 +00:00
parent f82d8064d4
commit dd847baadd
3 changed files with 25 additions and 1 deletions

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@ -10050,7 +10050,7 @@ Decl *Sema::ActOnFriendFunctionDecl(Scope *S, Declarator &D,
diag::err_friend_is_member);
DCScope = getScopeForDeclContext(S, DC);
// C++ [class.friend]p6:
// A function can be defined in a friend declaration of a class if and
// only if the class is a non-local class (9.8), the function name is
@ -10142,6 +10142,15 @@ Decl *Sema::ActOnFriendFunctionDecl(Scope *S, Declarator &D,
}
}
// FIXME: This is an egregious hack to cope with cases where the scope stack
// does not contain the declaration context, i.e., in an out-of-line
// definition of a class.
Scope FakeDCScope(S, Scope::DeclScope, Diags);
if (!DCScope) {
FakeDCScope.setEntity(DC);
DCScope = &FakeDCScope;
}
bool AddToScope = true;
NamedDecl *ND = ActOnFunctionDeclarator(DCScope, D, DC, TInfo, Previous,
move(TemplateParams), AddToScope);

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@ -4455,6 +4455,9 @@ Sema::TemplateParameterListsAreEqual(TemplateParameterList *New,
/// false. Otherwise, issues a diagnostic and returns true.
bool
Sema::CheckTemplateDeclScope(Scope *S, TemplateParameterList *TemplateParams) {
if (!S)
return false;
// Find the nearest enclosing declaration scope.
while ((S->getFlags() & Scope::DeclScope) == 0 ||
(S->getFlags() & Scope::TemplateParamScope) != 0)

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@ -0,0 +1,12 @@
// RUN: %clang_cc1 -fsyntax-only -verify %s
// <rdar://problem/10204947>
namespace N {
class X;
};
class N::X {
template<typename T> friend const T& f(const X&);
friend const int& g(const X&);
friend class Y;
};