How to Efficiently Convert Between a Byte and Eight Boolean Values?

Creating a Byte from Boolean Values and Vice Versa

Problem:

You have eight Boolean variables and desire to merge them into a single byte. Conversely, you aim to decode a byte into eight distinct Boolean values.

Solution:

Hard Way:

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;
}

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

Cool Way:

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;
};

Assign to one union member and read from another. Note that the order of bits in Bits is implementation-defined.

Caution:

Using unions in this manner is well-defined in ISO C99 and supported by some C compilers, but it is Undefined Behaviour in ISO C . Use memcpy or std::bit_cast for portable type-punning in C 20.

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