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c++: only declare satisfied friends
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A friend declaration can only have constraints if it is defined.  If
multiple instantiations of a class template define the same friend function
signature, it's an error, but that shouldn't happen if it's constrained to
only be declared in one instantiation.

Currently we don't mangle requirements, so the foos all mangle the same and
actually instantiating #1 will break, but for now we can test that they're
considered distinct.

gcc/cp/ChangeLog:

	* pt.cc (tsubst_friend_function): Check satisfaction.

gcc/testsuite/ChangeLog:

	* g++.dg/cpp2a/concepts-friend11.C: New test.
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jicama committed Nov 14, 2022
1 parent e7c12a9 commit c41bbfc
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3 changes: 3 additions & 0 deletions gcc/cp/pt.cc
Original file line number Diff line number Diff line change
Expand Up @@ -11284,6 +11284,9 @@ tsubst_friend_function (tree decl, tree args)
not_tmpl = DECL_TEMPLATE_RESULT (new_friend);
new_friend_result_template_info = DECL_TEMPLATE_INFO (not_tmpl);
}
else if (!constraints_satisfied_p (new_friend))
/* Only define a constrained hidden friend when satisfied. */
return error_mark_node;

/* Inside pushdecl_namespace_level, we will push into the
current namespace. However, the friend function should go
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21 changes: 21 additions & 0 deletions gcc/testsuite/g++.dg/cpp2a/concepts-friend11.C
Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
// CWG2596
// { dg-do compile { target c++20 } }

struct Base {};

int foo(Base&) { return 0; } // #0

template<int N>
struct S : Base {
friend int foo(Base&) requires (N == 1) { return 1; } // #1
// friend int foo(Base&) requires (N == 2) { return 3; } // #2
};

S<1> s1;
S<2> s2; // OK, no conflict between #1 and #0
int x = foo(s1); // { dg-error "ambiguous" }
int y = foo(s2); // OK, selects #0

// ??? currently the foos all mangle the same, so comment out #2
// and only test that #1 isn't multiply defined and overloads with #0.
// The 2596 example does not include #0 and expects both calls to work.

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