Fix merging of two arity-only pack deductions.

If we deduced the arity of a pack in two different ways, but didn't
deduce an element of the pack in either of those deductions, we'd merge
that element to produce a null template argument, which we'd incorrectly
interpret as the merge having failed.

Testcase based on one supplied by Hubert Tong.

Author: zygoloid

clang/lib/Sema/SemaTemplateDeduction.cpp

@@ -355,7 +355,7 @@ checkDeducedTemplateArguments(ASTContext &Context,
       TemplateArgument Merged = checkDeducedTemplateArguments(
           Context, DeducedTemplateArgument(*XA, X.wasDeducedFromArrayBound()),
           DeducedTemplateArgument(*YA, Y.wasDeducedFromArrayBound()));
-      if (Merged.isNull())
+      if (Merged.isNull() && !(XA->isNull() && YA->isNull()))
         return DeducedTemplateArgument();
       NewPack.push_back(Merged);
     }

clang/test/SemaTemplate/deduction.cpp

@@ -581,3 +581,19 @@ namespace PR44890 {
     return w.get<0>();
   }
 }
+
+namespace merge_size_only_deductions {
+#if __cplusplus >= 201703L
+  // Based on a testcase by Hubert Tong.
+  template<typename ...> struct X {};
+  template<auto ...> struct Y {};
+  template<typename T> struct id { using Type = T; };
+
+  template<typename ...T, typename T::Type ...V>
+    int f(X<char [V] ...>, Y<V ...>, X<T ...>);
+
+  using size_t = __SIZE_TYPE__;
+  int a = f(X<char [1], char [2]>(), Y<(size_t)1, (size_t)2>(), X<id<size_t>, id<size_t>>());
+  int b = f(X<char [1], char [2]>(), Y<1, 2>(), X<id<int>, id<int>>());
+#endif
+}