Programmer Humor

typedef struct {
    bool a: 1;
    bool b: 1;
    bool c: 1;
    bool d: 1;
    bool e: 1;
    bool f: 1;
    bool g: 1;
    bool h: 1;
} __attribute__((__packed__)) not_if_you_have_enough_booleans_t;

Depending on the language

I set all 8 bits to 1 because I want it to be really true.

string boolEnable = "True";

Then you need to ask yourself: Performance or memory efficiency? Is it worth the extra cycles and instructions to put 8 bools in one byte and & 0x bitmask the relevant one?

Wait till you find out about alignment and padding

Back in the day when it mattered, we did it like

#define BV00		(1 <<  0)
#define BV01		(1 <<  1)
#define BV02		(1 <<  2)
#define BV03		(1 <<  3)
...etc

#define IS_SET(flag, bit)	((flag) & (bit))
#define SET_BIT(var, bit)	((var) |= (bit))
#define REMOVE_BIT(var, bit)	((var) &= ~(bit))
#define TOGGLE_BIT(var, bit)	((var) ^= (bit))

....then...
#define MY_FIRST_BOOLEAN BV00
SET_BIT(myFlags, MY_FIRST_BOOLEAN)

The 8-bit Intel 8051 family provides a dedicated bit-addressable memory space (addresses 20h-2Fh in internal RAM), giving 128 directly addressable bits. Used them for years. I'd imagine many microcontrollers have bit-width variables.

bit myFlag = 0;

Or even return from a function:

bit isValidInput(unsigned char input) { // Returns true (1) if input is valid, false (0) otherwise return (input >= '0' && input <= '9'); }

In the industrial automation world and most of the IT industry, data is aligned to the nearest word. Depending on architecture, that's usually either 16, 32, or 64 bits. And that's the space a single Boolean takes.