How should I define a set number of bits in a typedef?

Question

The Problem

I'm currently trying to simulate some firmware in C++11. In the firmware we have a fixed data length of 32 bits, we split this 32 bits into smaller packets e.g we have a packet which as a size of 9 bits, another of 6 which gets packed into the 32 bit word.

In C++ I want to ensure the data I type in is of those lengths. I don't care if I overflow, just that only the 9 bits are operated on or passed onto another function.

Ideally I'd like some simple typedef like:

only_18_bits some_value;

My Attempt

struct sel_vals{
int_fast32_t m_val : 18;
int_fast8_t c_val : 5;
}

But this is a little annoying as I'd have to do this whenever I want to use it:

sel_vals somevals;
somevals.m_val = 5;

Seems a little verbose to me plus I have to declare the struct first.

Also for obvious reasons, I can't just do something like:

typedef sel_vals.m_val sel_vals_m_t;
typedef std::vector<sel_vals_m_t>;

I could use std::bitset<9> but whenever I want to do some maths I have to convert it to unsigned, it just gets a little messy. I want to avoid mess.

Any ideas?

Answer 1

I would suggest a wrapper facade, something along these lines:

#include <cstdint>

template<int nbits> class bits {

   uint64_t value;

   static const uint64_t mask = (~(uint64_t)0) >> (64-nbits);

public:
   bits(uint64_t initValue=0) : value(initValue & mask) {}

   bits &operator=(uint64_t newValue)
   {
       value=newValue & mask;
   }

   operator uint64_t() const { return value; }
};

//

bits<19> only_19_bits_of_precision;

With a little bit of work, you can define math operator overloads that directly operate on these templates.

With a little bit of more work, you could work this template to pick a smaller internal value, uint32_t, uint16_t, or uint8_t, if the nbits template parameter is small enough.