Any way to create a bitfield reference wrapper in a generic way?

Question

I am creating wrappers for C structs containing register definitions with heavy use of bitfields. I would like to create reference getters for all of them, to provide consistent shorthand API (real structures may use deep nesting with long names only a HW engineer would love).

For regular values, returning a reference is easy, but for bitfields, I had to create a wrapper similar to std::bitset::reference:

#include <cstdint>
#include <cstdio>

class storage {
    public:
    auto get_x() { return x_ref{ *this }; }
    auto get_y() { return y_ref{ *this }; }
    auto& get_z() { return z; }

    private:
    uint8_t x : 3;
    uint8_t y : 5;
    uint8_t z;

    struct x_ref {
        storage& store;
        operator uint8_t() { return store.x; }
        auto& operator=(uint8_t value) { store.x = value; return *this; }
    };
    
    struct y_ref {
        storage& store;
        operator uint8_t() { return store.y; }
        auto& operator=(uint8_t value) { store.y = value; return *this; }
    };
};

int main(int argc, char **argv) {
    storage s{};
    s.get_x() = 3;
    s.get_y() = 5;
    s.get_z() = 7;
    
    uint8_t x = s.get_x(), y = s.get_y(), z = s.get_z();
    std::printf("%u, %u\n", x, y, z);
}

Is there any way to create x/y_ref as a generic template? I would like to be able to write something like auto get_x() { return ref<&storage::x>{ *this }; } which is not a valid syntax for bitfields. std::bitset::reference does not have this problem, since it always stores a reference to uint8_t& (or larger word) and index, so it doesn't have anything to template on.

As you cannot use (member) references, remain lambda (but I fear you have to type almost the whole getter/setter) or MACRO. — Jarod42
– Jarod42, Commented Jun 24 at 8:35
OT: Instead of reimplementing the bitfield, use the existi8ng C structure itself. Bitfields are one of the worst defined features in both C and C++, and most things about them are implementation specific. — Some programmer dude
– Some programmer dude, Commented Jun 24 at 8:38
@Jarod42 By lambda you mean something like return ref{ [this]() { return this->x; }, [this](auto value) { this->x = value; } };? — Dominik Kaszewski
– Dominik Kaszewski, Commented Jun 24 at 8:48
@Someprogrammerdude I want to provide a shorthand for a subset of bitfields that application actually uses, which are hidden all over the place. s.get_foo() vs s.feature_x_static_values.u.version_2137.blah.blah.blah.finally_getting_somewhere.oh_here_it_is.my_foo, just with much more useless Hungarian. — Dominik Kaszewski
– Dominik Kaszewski, Commented Jun 24 at 8:51

Eljay · Accepted Answer · 2025-06-24 12:35:45Z

Here is one way to do it, but as Jarod42 said the lambdas are about as long as writing a struct helper.

I intentionally made the assignment return a void rather than the conventional return *this;. Chaining assignment is probably not the right thing to do here.

#include <cstdint>
#include <cstdio>

template <typename GET, typename SET>
struct var_ref {
    GET getter;
    SET setter;

    using store_t = decltype(getter());
    var_ref(GET g, SET s) : getter{g}, setter{s} {}

    // Using implicit conversion as the getter.
    operator store_t() { return getter(); }

    // Using void= as the setter.
    void operator=(store_t value) { setter(value); }
};

class storage {
    uint8_t x : 3;
    uint8_t y : 5;
    uint8_t z;

public:
    auto get_x() { return var_ref{[this]{ return x; }, [this](uint8_t value) { x = value; }}; }
    auto get_y() { return var_ref{[this]{ return y; }, [this](uint8_t value) { y = value; }}; }
    auto& get_z() { return z; }
};

int main() {
    storage s{};
    s.get_x() = 3;
    s.get_y() = 5;
    s.get_z() = 7;

    uint8_t x = s.get_x(), y = s.get_y(), z = s.get_z();
    std::printf("%u, %u, %u\n", x, y, z);
}

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