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You are saying that static_assert would not trigger here, correct? That makes sense to me, but it's not what is in the proposal [1] - see the "Not proposed" sec

by toth 10y ago

You are saying that static_assert would not trigger here, correct? That makes sense to me, but it's not what is in the proposal [1] - see the "Not proposed" section. [1] http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0292r1.html http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p029...

saynsedit 10y ago
No, I'm saying static_assert in non-instantiated templates would not trigger. template<class T> class foo; template<> class foo<int> { static_assert(false, "bad"); }; template<> class foo<float> {}; template<class T> foo<T> foo_fn(T v) { if constexpr (std::is_same<T, int>::value) { return foo<int>{}; } else if constexpr (std::is_same<T, float>::value) { return foo<float>{}; } else { return foo<T>{}; } } foo_fn(1.0); // <-- does not trigger static assert foo_fn(1); // <-- triggers static assert This is consistent with the wording of the proposal.
- Kristine1975 10y ago
  This does not compile: template<> class foo<int> { static_assert(false, "bad"); }; Reason: "false" is not a dependent name, so the compiler is allowed to check the static_assert when the template is being declared. (Of course since this is a full specialization, there are no dependent names.) See C++14 Standard §14.6.2 (Dependent Names) or http://en.cppreference.com/w/cpp/language/dependent_name http://en.cppreference.com/w/cpp/language/dependent_name (not sure how accurate that page is, though). P.S: It might compile with Microsoft's Visual Studio, but only because that compiler doesn't implement two-phase name lookup, effectively turning every name into a dependent name.
  - saynsedit 10y ago
    Oh tricky! I did write that code on my phone so hopefully you'll excuse the sloppiness. Thanks to HN I have miraculously learned a new thing about C++ when I was sure I knew it all; though, not surprised. Also this dependent name behavior of static_assert seems to be a precedent for always checking static_assert in constexpr if.

by toth 10y ago

saynsedit 10y ago

No, I'm saying static_assert in non-instantiated templates would not trigger. template<class T> class foo; template<> class foo<int> { static_assert(false, "bad"); }; template<> class foo<float> {}; template<class T> foo<T> foo_fn(T v) { if constexpr (std::is_same<T, int>::value) { return foo<int>{}; } else if constexpr (std::is_same<T, float>::value) { return foo<float>{}; } else { return foo<T>{}; } } foo_fn(1.0); // <-- does not trigger static assert foo_fn(1); // <-- triggers static assert This is consistent with the wording of the proposal.

Kristine1975 10y ago
This does not compile: template<> class foo<int> { static_assert(false, "bad"); }; Reason: "false" is not a dependent name, so the compiler is allowed to check the static_assert when the template is being declared. (Of course since this is a full specialization, there are no dependent names.) See C++14 Standard §14.6.2 (Dependent Names) or http://en.cppreference.com/w/cpp/language/dependent_name http://en.cppreference.com/w/cpp/language/dependent_name (not sure how accurate that page is, though). P.S: It might compile with Microsoft's Visual Studio, but only because that compiler doesn't implement two-phase name lookup, effectively turning every name into a dependent name.
- saynsedit 10y ago
  Oh tricky! I did write that code on my phone so hopefully you'll excuse the sloppiness. Thanks to HN I have miraculously learned a new thing about C++ when I was sure I knew it all; though, not surprised. Also this dependent name behavior of static_assert seems to be a precedent for always checking static_assert in constexpr if.