如何从8个bool值中创建一个字节(反之亦然)?

问题描述

我有8个bool变量，我想将它们合并"为一个字节.

I have 8 bool variables, and I want to "merge" them into a byte.

是否有一种简单/首选的方法来做到这一点?

Is there an easy/preferred method to do this?

反过来，将一个字节解码为8个独立的布尔值又如何呢?

How about the other way around, decoding a byte into 8 separate boolean values?

我假设这不是一个不合理的问题，但是由于我无法通过Google找到相关文档，因此这可能是您的直觉都不正确"的另一种情况.

I come in assuming it's not an unreasonable question, but since I couldn't find relevant documentation via Google, it's probably another one of those "nonono all your intuition is wrong" cases.

推荐答案

困难的方法:

unsigned char ToByte(bool b[8])
{
    unsigned char c = 0;
    for (int i=0; i < 8; ++i)
        if (b[i])
            c |= 1 << i;
    return c;
}

并且:

void FromByte(unsigned char c, bool b[8])
{
    for (int i=0; i < 8; ++i)
        b[i] = (c & (1<<i)) != 0;
}

或者很酷的方式:

struct Bits
{
    unsigned b0:1, b1:1, b2:1, b3:1, b4:1, b5:1, b6:1, b7:1;
};
union CBits
{
    Bits bits;
    unsigned char byte;
};

然后，您可以分配给工会的一个成员，并从另一个成员中读取.但是请注意，Bits中的位顺序是由实现定义的.

Then you can assign to one member of the union and read from another. But note that the order of the bits in Bits is implementation defined.

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