2位大小可变 [英] 2 bits size variable

问题描述

我需要定义一个结构，它具有大小2位和6位的数据成员。
我应该为每个成员使用 char 类型吗？或者，为了不浪费内存，我可以使用：2 \ ：6 符号？
我该怎么做呢？
我可以为2或6位类型定义一个typedef吗？

I need to define a struct which has data members of size 2 bits and 6 bits. Should I use char type for each member?Or ,in order not to waste a memory,can I use something like :2\ :6 notation? how can I do that? Can I define a typedef for 2 or 6 bits type?

推荐答案

>

You can use something like:

typedef struct {
    unsigned char SixBits:6;
    unsigned char TwoBits:2;
} tEightBits;

，然后使用：

tEightBits eight;
eight.SixBits = 31;
eight.TwoBits = 3;

但是，老实说，除非你必须遵守应用程序外部的打包数据，或者您处于非常内存限制的情况，这种内存节省通常不值得。如果不需要一直使用bitwise和bithift操作来打包和解压缩数据，您会发现代码的速度要快得多。

But, to be honest, unless you're having to comply with packed data external to your application, or you're in a very memory constrained situation, this sort of memory saving is not usually worth it. You'll find your code is a lot faster if it's not having to pack and unpack data all the time with bitwise and bitshift operations.

还要记住，使用除 _Bool ， signed int 或 unsigned int 是实现的问题。具体来说， unsigned char 可能无处不在。

Also keep in mind that use of any type other than _Bool, signed int or unsigned int is an issue for the implementation. Specifically, unsigned char may not work everywhere.

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