Bit manipulation

Learn bit manipulation patterns in embedded.

Bit manipulation

In embedded systems it is common to change specific bits of a register. A register is 32 bits; each bit controls a hardware feature. Bit manipulation sets, clears and toggles individual bits.

Three core operations: set (reg |= (1 << n)), clear (reg &= ~(1 << n)) and toggle (reg ^= (1 << n)). Here (1 << n) produces a mask: only the n-th bit is 1. |= sets a bit, &= ~ clears it, ^= toggles it. These operations are not atomic; interrupt safety requires care.

Masking and reading

To read a specific bit you AND with a mask: if (reg & (1 << n)). This gives true if the n-th bit is 1, false if 0. To read the lower part of a register: value = reg & 0xFF; (lower 8 bits).

To write multiple bits, first clear then set: reg &= ~0xF; reg |= (value & 0xF);. This replaces the lower 4 bits with the lower 4 bits of value. Masking is the standard way to manage sub-fields of hardware registers in embedded C. Every hardware driver uses these patterns.

Exercise Write a bit-set program

Define unsigned int reg = 0;. Set bit 0 with reg |= (1 << 0);. Set bit 2 with reg |= (1 << 2);. In int main(void) print with printf("Kayıt: %d\n", reg) (result should be 5: binary 101).

Show example solution

#include <stdio.h>

int main(void) {
    unsigned int reg = 0;
    reg |= (1 << 0);
    reg |= (1 << 2);
    printf("Kayıt: %d\n", reg);
    return 0;
}

Test yourself

Which operation sets a bit?
- reg &= ~(1 << n)
- reg = (1 << n)
- reg |= (1 << n)
- reg ^= (1 << n)
Show answer

reg |= (1 << n) — |= combines the mask with OR; the bit is set.

Complete the set-bit operation

reg |= (1 << ____);  /* bit 0'ı kur */

Show answer

reg |= (1 << 0);  /* bit 0'ı kur */

Match bit operation

Set

|= (1 << n)
Clear

&= ~(1 << n)
Toggle

^= (1 << n)
Read

& (1 << n)

Next lesson: Hardware timing

We Are Enhancing Your Experience

Cookie Preferences

Strictly Necessary Cookies

Analytical Cookies