如何使用朋友访问STL私人成员

Question

When I do this...

auto t = typeid(float);

... I get:

Error C2248 'type_info::type_info': cannot access private member declared in class 'type_info'

Of course, I know I can do this...

auto t = typeid(float).name();

But if I wanted to use the first expression, how would I go about tampering with the type_info class and using friend to achieve my objective? Any suggestions?

Answer 1

You can't make your class a friend of a standard class.

A typeid expression is an lvalue, so you can simply bind a reference to it:

auto& t = typeid(float);

Answer 2

The reason you cannot do this is the copy constructor and the assignment operator of std::type_info are marked as delete per [type.info]. So

auto t = typeid(float);

Will attempt to call the copy constructor which will fail.

As TartanLlama points out in his answer you can capture a reference to it with

auto& t = typeid(float);

Answer 3

 Error C2248 'type_info::type_info': cannot access private member declared in class 'type_info' 

This error message refers to the copy constructor of type_info which you've tried to invoke with auto t = typeid(float); .

But if I wanted to use the first expression, how would I go about tampering with the type_info class and using friend to achieve my objective? Any suggestions?

Even if you could manipulate type_info 's definition, which you cannot , what would you possibly do? The copy constructor is not private because it needs to be hidden from you but because the language does not define what it means to copy a type_info . In fact, even if you somehow declared the member function as public (which is already purely hypothetical), there would be no definition and thus a linker error.

I don't know why GCC is not more specific in its error message here. With C++11, the copy constructor is not just private but deleted and that's the reason it cannot be used. Coincidentally, MSVC's error message is more appropriate, as it says:

error C2280: 'type_info::type_info(const type_info &)' : 
attempting to reference a deleted function

---

A modern C++11 solution to your problem is to use std::type_index , which is a copyable wrapper around std::type_info . Here is an example:

#include <typeindex>

int main()
{
    auto t = std::type_index(typeid(float));
}

Answer 4

Class std::type_info has deleted copy constructor.

type_info(const type_info& rhs) = delete; // cannot be copied
                                  ^^^^^^^^ 

On the other hand (5.2.8 Type identification)

1 The result of a typeid expression is an lvalue of static type const std::type_info (18.7.1) and dynamic type const std::type_info or const name where name is an implementation-defined class publicly derived from std :: type_info which preserves the behavior described in 18.7.1.69 The lifetime of the object referred to by the lvalue extends to the end of the program . Whether or not the destructor is called for the std::type_info object at the end of the program is unspecified.

So you can write for example

#include <iostream>

int main()
{
    decltype( auto ) t = typeid( float );

    std::cout << t.name() << std::endl;

    return 0;
}