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

Question

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.

Pointer-to-member. template <typename T, typename U, U T::*ptm>. Demo With a sufficiently recent compiler, can be further simplified to this — Igor Tandetnik
– Igor Tandetnik, Commented Aug 12, 2024 at 16:39
std::bind_front(&A::b) will give you a functor you can call with an A. Is this enough? — Wutz
– Wutz, Commented Aug 12, 2024 at 20:16

Richard · Accepted Answer · 2024-08-12 17:09:17Z

1

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);
}

answered Aug 12, 2024 at 17:09

Richard

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1 Comment

LibrarristShalinward Over a year ago

Marvelous! That's just what I want. Great Thanks! I've also thought about pointers but I didn't know it can be used like this.

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