How to use attribute (not its value) of a struct/class as template/function parameter?

To be short, I want to achieve something like "templatize" the operation of "getting an attribute of a object" as a generalized template function, like the function foo below.

template <
    typename T, 
    // something like
    "attribute" obj // or T::obj
>
"sth refer to type of obj" foo(T t)
{
    return t.obj; 
}

What I hope to achieve is, with any simple struct/class like

struct A
{
    int b; 
    type_of_c c; 
    ...
} a; // for any class/struct A with (non-function) attribute b in any type

foo<A, A::b>(a) may return a.b, and foo<A, A::c>(a) may return a.c.

So my question is, how can I express T::obj to set the information of "specifying attribute obj of class T" as the parameter of template?

Some limitations in my programming:

• Methods that unfriendly to compile and IDE auto-hints are not encouraged. e.g. sending attribute name in form of string instead of attribute itself, or involving macro.
• Add type of T::obj to the template is acceptable, but not encouraged.
• Tricks in any version of c++ is acceptable. Using newly published versions like C++20 is definitely ok.
• The solution is expected to work also for class template, if possible.

You can make use of pointer to members. To avoid having three template parameter, you can also make use of template argument deduction as shown below.

template < typename T, typename U>
U foo2(U (T::*obj), T t)
{
    return (t.*obj); 
}
int main()
{
    std::cout << foo2(&A::b, a);
}

Working demo

Solution 2

If you're okay with passing three template argument you can also do:

template < typename T, typename U, U (T::*obj)>
U foo(T t)
{
    return (t.*obj); 
}
int main()
{
    std::cout << foo<A, int, &A::b>(a);
}