Specifying bit size of array elements in a struct - c++

Question

Now I have a struct looking like this:
struct Struct {
uint8_t val1 : 2;
uint8_t val2 : 2;
uint8_t val3 : 2;
uint8_t val4 : 2;
} __attribute__((packed));
Is there a way to make all the vals a single array? The point is not space taken, but the location of all the values: I need them to be in memory without padding, and each occupying 2 bits. It's not important to have array, any other data structure with simple access by index will be ok, and not matter if it's plain C or C++. Read/write performance is important - it should be same (similar to) as simple bit operations, which are used now for indexed access.
Update:
What I want exactly can be described as
struct Struct {
uint8_t val[4] : 2;
} __attribute__((packed));

Answer 1

No, C only supports bitfields as structure members, and you cannot have arrays of them. I don't think you can do:
struct twobit {
uint8_t val : 2;
} __attribute__((packed));
and then do:
struct twobit array[32];
and expect array to consist of 32 2-bit integers, i.e. 8 bytes. A single char in memory cannot contain parts of different structs, I think. I don't have the paragraph and verse handy right now though.
You're going to have to do it yourself, typically using macros and/or inline functions to do the indexing.

Answer 2

You have to manually do the bit stuff that's going on right now:
constexpr uint8_t get_mask(const uint8_t n)
{
return ~(((uint8_t)0x3)<<(2*n));
}
struct Struct2
{
uint8_t val;
inline void set_val(uint8_t v,uint8_t n)
{
val = (val&get_mask(n))|(v<<(2*n));
}
inline uint8_t get_val(uint8_t n)
{
return (val&~get_mask(n))>>(2*n);
}
//note, return type only, assignment WONT work.
inline uint8_t operator[](uint8_t n)
{
return get_val(n);
}
};
Note that you may be able to get better performance if you use actual assembly commands.
Also note that, (almost) no matter what, a uint8_t [4] will have better performance than this, and a processor aligned type (uint32_t) may have even better performance.