具有混叠的可变大小的位域 [英] Variable-sized bitfields with aliasing

问题描述

我有一些结构包含一个位域，其大小可能不同。示例：

I have some struct containig a bitfield, which may vary in size. Example:

struct BitfieldSmallBase {
    uint8_t a:2;
    uint8_t b:3;
    ....
}

struct BitfieldLargeBase {
    uint8_t a:4;
    uint8_t b:5;
    ....
}

strong>一次访问所有位：

and a union to access all bits at once:

template<typename T>
union Bitfield 
{
    T bits;
    uint8_t all;    // <-------------  Here is the problem

    bool operator & (Bitfield<T> x) const {
        return !!(all & x.all);
    }
    Bitfield<T> operator + (Bitfield<T> x) const {
        Bitfield<T> temp;
        temp.all = all + x.all;   //works, because I can assume no overflow will happen
        return temp;
    }
    ....
}

typedef Bitfield<BitfieldSmallBase> BitfieldSmall;
typedef Bitfield<BitfieldLargeBase> BitfieldLarge;

问题是：对于一些位域基类，uint8_t是不够的。 BitfieldSmall适合一个uint8_t，但BitfieldLarge不。数据需要尽可能紧密地打包（它将由SSE指令后面处理），所以总是使用uint16_t是没有问题的。有没有办法用一个整数类型声明所有字段，其大小与位字段相同？还是另一种方式来访问位作为一个整体？

The problem is: For some bitfield base classes, an uint8_t is not sufficient. BitfieldSmall does fit into a uint8_t, but BitfieldLarge does not. The data needs to be packed as tightly as possible (it will be handled by SSE instructions later), so always using uint16_t is out of question. Is there a way to declare the "all" field with an integral type, whose size is the same as the bitfield? Or another way to access bits as a whole?

我当然可以放弃使用模板，并明确声明每种位域，但我想避免代码重复（有很多操作符和成员函数的列表）。

I can of course forego the use of the template and declare every kind of bitfield explicitly, but I would like to avoid code repetition (there is quite a list of operators und member functions).

推荐答案

您可以使整型类型为模板参数。

You could make the integral type a template parameter as well.

template<typename T, typename U>
union Bitfield 
{
    T bits;
    U all;
}

typedef Bitfield<BitfieldSmallBase, uint8_t>  BitfieldSmall;
typedef Bitfield<BitfieldLargeBase, uint16_t> BitfieldLarge;

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