When transforming a substituted template type parameter, try to
transform the type that replaces the template type parameter. In the
vast majority of cases, there's nothing to do, because most template
type parameters are replaced with something non-dependent that doesn't
need further transformation. However, when we're dealing with the
default template arguments of template template parameters, we might
end up replacing a template parameter (of the template template
parameter) with a template parameter of the enclosing template.
This addresses part of PR9016, but not within function
templates. That's a separate issue.
llvm-svn: 127091
diff --git a/clang/test/SemaTemplate/issue150.cpp b/clang/test/SemaTemplate/issue150.cpp
index 0d79307..2cfa8c5 100644
--- a/clang/test/SemaTemplate/issue150.cpp
+++ b/clang/test/SemaTemplate/issue150.cpp
@@ -69,3 +69,29 @@
int check0[is_same<X<int, int, Y>::type, Y<int, int*, int* const> >::value? 1 : -1];
}
+
+namespace PR9016 {
+ template<typename > struct allocator ;
+ template<typename > struct less ;
+
+ template<class T, template<class> class Compare, class Default,
+ template<class> class Alloc>
+ struct interval_set { };
+
+ template <class X, template<class> class = less> struct interval_type_default {
+ typedef X type;
+ };
+
+ template <class T,
+ template<class _T, template<class> class Compare = less,
+ class = typename interval_type_default<_T,Compare>::type,
+ template<class> class = allocator> class IntervalSet>
+ struct ZZZ
+ {
+ IntervalSet<T> IntervalSetT;
+ };
+
+ void test() {
+ ZZZ<int, interval_set> zzz;
+ }
+}