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      std::is_convertible,std::is_nothrow_convertible

      From cppreference.com
      <cpp‎ |types
       
       
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      Type properties
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      Compile-time rational arithmetic
      Compile-time integer sequences
       
      Defined in header<type_traits>
      template<class From,class To>
      struct is_convertible;
      (1)(since C++11)
      template<class From,class To>
      struct is_nothrow_convertible;
      (2)(since C++20)
      1) If the imaginary function definitionTo test(){returnstd::declval<From>();} is well-formed, (that is, eitherstd::declval<From>() can be converted toTo usingimplicit conversions, or bothFrom andTo are possibly cv-qualifiedvoid), provides the member constantvalue equal totrue. Otherwisevalue isfalse. For the purposes of this check, the use ofstd::declval in the return statement is not considered anODR-use.

      IfTo is a reference type and atemporary object would be created when bindingstd::declval<From>() toTo, thereturn statement in the imaginary function is considered well-formed, even though such binding is ill-formed in an actual function.

      (since C++26)
      Access checks are performed as if from a context unrelated to either type. Only the validity of the immediate context of the expression in the return statement (including conversions to the return type) is considered.
      2) Same as(1), but the conversion is alsonoexcept.

      IfFrom orTo is not a complete type, (possibly cv-qualified)void, or an array of unknown bound, the behavior is undefined.

      If an instantiation of a template above depends, directly or indirectly, on an incomplete type, and that instantiation could yield a different result if that type were hypothetically completed, the behavior is undefined.

      If the program adds specializations for any of the templates described on this page, the behavior is undefined.

      Contents

      [edit]Helper variable template

      template<class From,class To>
      constexprbool is_convertible_v= is_convertible<From, To>::value;
      (since C++17)
      template<class From,class To>
      constexprbool is_nothrow_convertible_v= is_nothrow_convertible<From, To>::value;
      (since C++20)
      [edit]

      Inherited fromstd::integral_constant

      Member constants

      value
      [static]
      true ifFrom is convertible toTo,false otherwise
      (public static member constant)

      Member functions

      operator bool
      converts the object tobool, returnsvalue
      (public member function)
      operator()
      (C++14)
      returnsvalue
      (public member function)

      Member types

      Type Definition
      value_typebool
      typestd::integral_constant<bool, value>

      [edit]Possible implementation

      is_convertible (1)
      namespace detail{template<class T>auto test_returnable(int)-> decltype(void(static_cast<T(*)()>(nullptr)),std::true_type{});template<class>auto test_returnable(...)->std::false_type; template<class From,class To>auto test_implicitly_convertible(int)-> decltype(void(std::declval<void(&)(To)>()(std::declval<From>())),std::true_type{});template<class,class>auto test_implicitly_convertible(...)->std::false_type;}// namespace detail template<class From,class To>struct is_convertible:std::integral_constant<bool,(decltype(detail::test_returnable<To>(0))::value&&     decltype(detail::test_implicitly_convertible<From, To>(0))::value)||(std::is_void<From>::value&&std::is_void<To>::value)>{};
      is_nothrow_convertible (2)
      template<class From,class To>struct is_nothrow_convertible:std::conjunction<std::is_void<From>,std::is_void<To>>{}; template<class From,class To>    requires        requires{static_cast<To(*)()>(nullptr);{std::declval<void(&)(To)noexcept>()(std::declval<From>())}noexcept;}struct is_nothrow_convertible<From, To>:std::true_type{};

      [edit]Notes

      Gives well-defined results for reference types, void types, array types, and function types.

      Currently the standard has not specified whether the destruction of the object produced by the conversion (either a result object or a temporary bound to a reference) is considered as a part of the conversion. This isLWG issue 3400.

      All known implementations treat the destruction as a part of the conversion, as proposed inP0758R1.

      Feature-test macroValueStdFeature
      __cpp_lib_is_nothrow_convertible201806L(C++20)std::is_nothrow_convertible

      [edit]Example

      Run this code
      #include <iomanip>#include <iostream>#include <string>#include <string_view>#include <type_traits> class E{public:template<class T> E(T&&){}}; int main(){class A{};class B:public A{};class C{};class D{public: operator C(){return c;} C c;};     static_assert(std::is_convertible_v<B*, A*>);    static_assert(!std::is_convertible_v<A*, B*>);    static_assert(std::is_convertible_v<D, C>);    static_assert(!std::is_convertible_v<B*, C*>);// Note that the Perfect Forwarding constructor makes the class E be// "convertible" from everything. So, A is replaceable by B, C, D..:    static_assert(std::is_convertible_v<A, E>);     static_assert(!std::is_convertible_v<std::string_view,std::string>);    static_assert(std::is_convertible_v<std::string,std::string_view>); auto stringify=[]<typename T>(T x){ifconstexpr(std::is_convertible_v<T,std::string> or                      std::is_convertible_v<T,std::string_view>)return x;elsereturnstd::to_string(x);}; using std::operator"" s, std::operator"" sv;constchar* three="three"; std::cout<<std::quoted(stringify("one"s))<<' '<<std::quoted(stringify("two"sv))<<' '<<std::quoted(stringify(three))<<' '<<std::quoted(stringify(42))<<' '<<std::quoted(stringify(42.0))<<'\n';}

      Output:

      "one" "two" "three" "42" "42.000000"

      [edit]See also

      (C++11)
      checks if a type is a base of the other type
      (class template)[edit]
      checks if a type is apointer-interconvertible (initial) base of another type
      (class template)[edit]
      checks if objects of a type arepointer-interconvertible with the specified subobject of that type
      (function template)[edit]
      specifies that a type is implicitly convertible to another type
      (concept)[edit]
      Retrieved from "https://en.cppreference.com/mwiki/index.php?title=cpp/types/is_convertible&oldid=176476"

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