How to create a byte out of 8 bool values (and vice versa)?

Question

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

Is there an easy/preferred method to do this?

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

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.

Answer 1

The 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; } 

And:

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

Or the 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; }; 

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.

Note that reading one union member after writing another is well-defined in ISO C99, and as an extension in several major C++ implementations (including MSVC and GNU-compatible C++ compilers), but is Undefined Behaviour in ISO C++. memcpy or C++20 std::bit_cast are the safe ways to type-pun in portable C++.

(Also, the bit-order of bitfields within a char is implementation defined, as is possible padding between bitfield members.)