std::is_convertible

std::is_convertible

template< class From, class To >
struct is_convertible;

If the imaginary function definition To test() { return std::declval<From>(); } is well-formed, (that is, either std::declval<From>() can be converted to To using implicit conversion, or both From and To are possibly cv-qualified void), provides the member constant value equal to true. Otherwise value is false. For the purposes of this check, the use of std::declval is in the return statement is not considered an odr-use.

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.

From and To shall each be a complete type, (possibly cv-qualified) void, or an array of unknown bound. Otherwise, the behavior is undefined.

Helper variable template

template< class From, class To >
inline constexpr bool is_convertible_v = is_convertible<From, To>::value;

Notes

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

Example

#include <iostream>
#include <type_traits>
 
int main() 
{
    class A {};
    class B : public A {};
    class C {};
 
    bool b2a = std::is_convertible<B*, A*>::value;
    bool a2b = std::is_convertible<A*, B*>::value;
    bool b2c = std::is_convertible<B*, C*>::value;
 
    std::cout << std::boolalpha;
    std::cout << b2a << '\n';
    std::cout << a2b << '\n';
    std::cout << b2c << '\n';
}

true
false
false

See also