c++17

问题 This code compiles: struct Info { constexpr Info(bool val) : counted(false), value(unsigned(val)) {} constexpr Info(unsigned val) : counted(true), value(val) {} bool counted; unsigned value; }; constexpr const auto data = std::array{ Info{true}, Info{42u} }; struct Foo { constexpr static inline const auto data = std::array{ Info{true}, Info{42u} }; }; This code does not: struct Foo { struct Info { constexpr Info(bool val) : counted(false), value(unsigned(val)) {} constexpr Info(unsigned val)

问题 I wanted to try out the new filesystem library in C++17, so tried copying the std::filesystem::current_path example from cppreference.com and compiling it using the latest version (16.0) of the MinGW distribution from nuwen.net on my Windows 10 x64 machine. This includes gcc v8.1 which should support the filesystem library according to cppreference.com compiler support page. Here is the code that I am trying to compile: #include <iostream> #include <filesystem> namespace fs = std::filesystem;

问题 I wanted to try out the new filesystem library in C++17, so tried copying the std::filesystem::current_path example from cppreference.com and compiling it using the latest version (16.0) of the MinGW distribution from nuwen.net on my Windows 10 x64 machine. This includes gcc v8.1 which should support the filesystem library according to cppreference.com compiler support page. Here is the code that I am trying to compile: #include <iostream> #include <filesystem> namespace fs = std::filesystem;

问题 Say I have [[nodiscard]] int foo () { return 0; } int main () { foo (); } then error: ignoring return value of ‘int foo()’, declared with attribute nodiscard [-Werror=unused-result] but if int x = foo (); then error: unused variable ‘x’ [-Werror=unused-variable] Is there a clean way of telling the compiler "I want to discard this [[nodiscard]] value"? 回答1: The WG14 nodiscard proposal discusses the rationale for allowing the diagnostic to be silenced by casting to void. It says casting to void

问题 Say I have [[nodiscard]] int foo () { return 0; } int main () { foo (); } then error: ignoring return value of ‘int foo()’, declared with attribute nodiscard [-Werror=unused-result] but if int x = foo (); then error: unused variable ‘x’ [-Werror=unused-variable] Is there a clean way of telling the compiler "I want to discard this [[nodiscard]] value"? 回答1: The WG14 nodiscard proposal discusses the rationale for allowing the diagnostic to be silenced by casting to void. It says casting to void

问题 I have set the following in my CMakeLists.txt: set (CMAKE_CXX_STANDARD 17) set (CMAKE_CXX_STANDARD_REQUIRED ON) set (CMAKE_CXX_EXTENSIONS OFF) However, CMake still allows g++ 6, even though it doesn't fully support c++17 (it has a c++1z standard, but not a c++17 standard). Is there a way to tell CMake to only allow compilers that fully support the standard and not just pieces of it? FWIW, I also tried setting cxx_relaxed_constexpr, which I think should have been the relevant language feature,

问题 I want to write a helper template that checks if a template parameter pack has a common type, i.e., if applying std::common_type to the pack defines a type. Using std::void_t with SFINAE I came up with the following definition: template<typename... Types, typename Enable = void> struct has_common_type : std::false_type { }; template<typename... Types> struct has_common_type<Types..., std::void_t<std::common_type<Types...>::type>> : std::true_type { }; This however does not work, because the

问题 I want to write a helper template that checks if a template parameter pack has a common type, i.e., if applying std::common_type to the pack defines a type. Using std::void_t with SFINAE I came up with the following definition: template<typename... Types, typename Enable = void> struct has_common_type : std::false_type { }; template<typename... Types> struct has_common_type<Types..., std::void_t<std::common_type<Types...>::type>> : std::true_type { }; This however does not work, because the

问题 I want to write a helper template that checks if a template parameter pack has a common type, i.e., if applying std::common_type to the pack defines a type. Using std::void_t with SFINAE I came up with the following definition: template<typename... Types, typename Enable = void> struct has_common_type : std::false_type { }; template<typename... Types> struct has_common_type<Types..., std::void_t<std::common_type<Types...>::type>> : std::true_type { }; This however does not work, because the

问题 The draft n4659 for C++17 describes the general principes of the language in chapter 4. In chapter 4.5, The C++ object model [intro.object], I cannot understand the meaning of one sentence (emphasize mine) 3 If a complete object is created (8.3.4) in storage associated with another object e of type “array of N unsigned char” or of type “array of N std::byte” (21.2.1), that array provides storage for the created object if: (3.1) — the lifetime of e has begun and not ended, and (3.2) — the