: false std::is_arithmetic<std::complex<double>>::value ?: false std::is_arithmetic_v<char> ?: true std::is_arithmetic_v<float> ?: true std::is_arithmetic_v<float*> ?: false std::is_arithmetic_v<std::complex<do
is_arithmetic_v<int const> == true and std::is_arithmetic_v<float> == true and std::is_arithmetic_v<float const> == true and std::is_arithmetic_v<std::size_t> == true and std::is_arithmetic_v<char&> == false and std::is_arithmetic_v<char*> == false and std::is_...
is_arithmetic_v<> (C++17 起)is_array<> (C++11 起)is_array_v<> (C++17 起)is_assignable<> (C++11 起)is_assignable_v<> (C++17 起)is_base_of<> (C++11 起)is_base_of_v<> (C++17 起)is_bind_expression<> (C++11 起)is_bind_expression_v<> (C++17 起)is_bounded_array<> (...
std::is_arithmetic 定义于头文件<type_traits> template<classT> structis_arithmetic; (C++11 起) 若T为算术类型(即整数类型或浮点类型)或其cv限定版本,则提供等于true的成员常量value。对于任何其他类型,value为false。 添加is_arithmetic或is_arithmetic_v(C++17 起)的特化的程序行为未定义。
inline constexpr bool is_reference_v = false; template <typename _Tp> inline constexpr bool is_reference_v<_Tp&> = true; template <typename _Tp> inline constexpr bool is_reference_v<_Tp&&> = true; #endif template <typename _Tp> inline constexpr bool is_arithmetic_v = is_arithmetic<...
is_integer identifies integer types (public static member constant of std::numeric_limits<T>) [static] is_floating_point checks if a type is a floating-point type (class template) (C++11) is_arithmetic checks if a type is an arithmetic type (class template) (C++11) is_enum che...
; } emm,相信你也看出问题所在了,当然了,编译器也会提示如下错误: error: no viable conversion from 'const int' to 'std::string' (aka...::false_type {}; template struct is_stringstd::enabl...
If std::is_arithmetic<T>::value is true, provides the member constant value equal to T(0) < T(-1). Otherwise, provides the member constant value equal to false. If the program adds specializations for std::is_unsigned or std::is_unsigned_v, the behavior is undefined. Template pa...
std::is_scalar 是一元类型特征 (UnaryTypeTrait) 。 如果T 是标量类型,那么提供的成员常量 value 等于true。对于其它任何类型,value 等于false。 如果程序添加了 std::is_scalar 或std::is_scalar_v 的特化,那么行为未定义。 模板形参T - 要检查的类型 ...
If T is a fundamental type (that is, arithmetic type, void, or nullptr_t), provides the member constant value equal true. For any other type, value is false. If the program adds specializations for std::is_fundamental or std::is_fundamental_v, the behavior is undefined. Template para...