On 3/11/20 2:10 PM, Jason Merrill wrote:
On 3/11/20 12:57 PM, Martin Sebor wrote:
On 3/9/20 6:08 PM, Jason Merrill wrote:
On 3/9/20 5:39 PM, Martin Sebor wrote:
On 3/9/20 1:40 PM, Jason Merrill wrote:
On 3/9/20 12:31 PM, Martin Sebor wrote:
On 2/28/20 1:24 PM, Jason Merrill wrote:
On 2/28/20 12:45 PM, Martin Sebor wrote:
On 2/28/20 9:58 AM, Jason Merrill wrote:
On 2/24/20 6:58 PM, Martin Sebor wrote:
-Wredundant-tags doesn't consider type declarations that are also
the first uses of the type, such as in 'void f (struct S);' and
issues false positives for those.  According to the reported that's
making it harder to use the warning to clean up LibreOffice.

The attached patch extends -Wredundant-tags to avoid these false
positives by relying on the same class_decl_loc_t::class2loc mapping as -Wmismatched-tags.  The patch also somewhat improves the detection of both issues in template declarations (though more work is still
needed there).

+         a new entry for it and return unless it's a declaration
+         involving a template that may need to be diagnosed by
+         -Wredundant-tags.  */
       *rdl = class_decl_loc_t (class_key, false, def_p);
-      return;
+      if (TREE_CODE (decl) != TEMPLATE_DECL)
+        return;

How can the first appearance of a class template be redundant?

I'm not sure I correctly understand the question.  The comment says
"involving a template" (i.e., not one of the first declaration of
a template).  The test case that corresponds to this test is:

   template <class> struct S7 { };
   struct S7<void> s7v;  // { dg-warning "\\\[-Wredundant-tags" }

where DECL is the TEPLATE_DECL of S7<void>.

As I mentioned, more work is still needed to handle templates right
because some redundant tags are still not diagnosed.  For example:

   template <class> struct S7 { };
   template <class T>
   using U = struct S7<T>;   // missing warning

When we get here for an instance of a template, it doesn't make sense to treat it as a new type.

If decl is a template and type_decl is an instance of that template, do we want to (before the lookup) change type_decl to the template or the corresponding generic TYPE_DECL, which should already be in the table?

I'm struggling with how to do this.  Given type (a RECORD_TYPE) and
type_decl (a TEMPLATE_DECL) representing the use of a template, how
do I get the corresponding template (or its explicit or partial
specialization) in the three cases below?

   1) Instance of the primary:
      template <class> class A;
      struct A<int> a;

   2) Instance of an explicit specialization:
      template <class> class B;
      template <> struct B<int>;
      class B<int> b;

   3) Instance of a partial specialization:
      template <class> class C;
      template <class T> struct C<T*>;
      class C<int*> c;

By trial and (lots of) error I figured out that in both (1) and (2),
but not in (3), TYPE_MAIN_DECL (TYPE_TI_TEMPLATE (type)) returns
the template's type_decl.

Is there some function to call to get it in (3), or even better,
in all three cases?

I think you're looking for most_general_template.

I don't think that's quite what I'm looking for.  At least it doesn't
return the template or its specialization in all three cases above.

Ah, true, that function stops at specializations.  Oddly, I don't think there's currently a similar function that looks through them. You could create one that does a simple loop through DECL_TI_TEMPLATE like is_specialization_of.

Thanks for the tip.  Even with that I'm having trouble with partial
specializations.  For example in:

   template <class>   struct S;
   template <class T> class S<const T>;
   extern class  S<const int> s1;
   extern struct S<const int> s2;  // expect -Wmismatched-tags

how do I find the declaration of the partial specialization when given
the type in the extern declaration?  A loop in my find_template_for()
function (similar to is_specialization_of) only visits the implicit
specialization S<const int> (i.e., its own type) and the primary.

Is that a problem?  The name is from the primary template, so does it matter for this warning whether there's an explicit specialization involved?

I don't understand the question.  S<const int> is an instance of
the partial specialization.  To diagnose the right mismatch the warning
needs to know how to find the template (i.e., either the primary, or
the explicit or partial specialization) the instance corresponds to and
the class-key it was declared with.  As it is, while GCC does diagnose
the right declaration (that of s2), it does that thanks to a bug:
because it finds and uses the type and class-key used to declare s1.
If we get rid of s1 it doesn't diagnose anything.

I tried using DECL_TEMPLATE_SPECIALIZATIONS() to get the list of
the partial specializations but it doesn't like any of the arguments
I've given it (it ICEs).

Martin

PS As an aside, both Clang and VC++ have trouble with templates as
well, just slightly different kinds of trouble.   Clang diagnoses
the declaration of the partial specialization while VC++ diagnoses
s1.  In other similar cases like the one below VC++ does the right
thing and Clang is silent.

  template <class>   struct S { };
  template <class T> class S<const T> { };

  template <class T> void f (class S<T>);          // VC++ warns
  template <class T> void g (struct S<const T>);   // GCC & VC++ warn


Martin


In (2) and (3) it won't distinguish between specializations of B or
C on different types.  In (2), the function returns the same result
for both:

   template <> struct B<int>;
   template <> struct B<char>;

In (3), it similarly returns the same result for both of

   template <class T> struct C<T*>;
   template <class T> struct C<const T>;

even though they are declarations of distinct types.


Jason




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