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# std::has_virtual_destructor

From cppreference.com
< cpp‎ | types

Defined in header `<type_traits>` |  |
template< class T > struct has_virtual_destructor; |  | (since C++11)

If `T` is a type with a virtual destructor, provides the member constant `value` equal true. For any other type, `value` is false.

## Inherited from std::integral_constant

### Member constants

value [static] | true if `T` has a virtual destructor , false otherwise (public static member constant)

### Member functions

operator bool | converts the object to bool, returns `value` (public member function)

### Member types

Type | Definition
`value_type` | `bool`
`type` | std::integral_constant<bool, value>

### [edit] Notes

If a class has a public virtual destructor, it can be derived from, and the derived object can be safely deleted through a pointer to the base object (C++FAQ Lite 20.7)

### [edit] Example

Run this code

    #include <iostream>
    #include <type_traits>
    #include <string>
    #include <stdexcept>

    int main()
    {
        std::cout << std::boolalpha
                  << "std::string has a virtual destructor? "
                  << std::has_virtual_destructor<std::string>::value << '\n'
                  << "std::runtime_error has a virtual destructor? "
                  << std::has_virtual_destructor<std::runtime_error>::value << '\n';
    }

Output:

    std::string has a virtual destructor? false
    std::runtime_error has a virtual destructor? true

### [edit] See also

is_destructible is_trivially_destructible is_nothrow_destructible (C++11)(C++11)(C++11) | checks if a type has a non-deleted destructor (class template) [edit]

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