c++: Add test for c++/93807
This PR was initially accepts-invalid, but I think it's actually valid C++20 code. My reasoning is that in C++20 we no longer require the declaration of operator== (#if-defed in the test), because C++20's [temp.names]/2 says "A name is also considered to refer to a template if it is an unqualified-id followed by a < and name lookup either finds one or more functions or finds nothing." so when we're parsing constexpr friend bool operator==<T>(T lhs, const Foo& rhs); we treat "operator==" as a template name, because name lookup of "operator==" found nothing and we have an operator-function-id, which is an unqualified-id, and it's followed by a <. So the declaration isn't needed to treat "operator==<T>" as a template-id. PR c++/93807 * g++.dg/cpp2a/fn-template20.C: New test.
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2020-04-22 Marek Polacek <polacek@redhat.com>
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PR c++/93807
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* g++.dg/cpp2a/fn-template20.C: New test.
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2020-04-22 Duan bo <duanbo3@huawei.com>
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PR testsuite/94712
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34
gcc/testsuite/g++.dg/cpp2a/fn-template20.C
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gcc/testsuite/g++.dg/cpp2a/fn-template20.C
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// PR c++/93807
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// { dg-do compile { target c++11 } }
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// In C++17, we need the following declaration to treat operator== as
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// a template name. In C++20, this is handled by [temp.names]/2.
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#if __cplusplus <= 201703L
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template <typename T>
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class Foo;
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template <typename T>
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constexpr bool operator==(T lhs, const Foo<T>& rhs);
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#endif
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template <typename T>
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class Foo {
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public:
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constexpr Foo(T k) : mK(k) {}
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constexpr friend bool operator==<T>(T lhs, const Foo& rhs);
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private:
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T mK;
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};
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template <typename T>
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constexpr bool
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operator==(T lhs, const Foo<T>& rhs)
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{
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return lhs == rhs.mK;
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}
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int
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main ()
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{
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return 1 == Foo<int>(1) ? 0 : 1;
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}
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