How can I toggle certain bits in an 8bit register?

Question

Say I have bits 0-3 I want to toggle given a certain register value, how can I do that?

eg:

unsigned char regVal = 0xB5; //1011 0101

// Toggle bits 0-3 (0101) to 1 without changing 4-7 where result should be 1011 1111

unsigned char result = regVal & (0x01 | 0x02 | 0x03 | 0x04);

or

unsigned char regVal = 0x6D; //0110 1101

// Toggle bits 4 and 7 to 1 without changing 1,2,3,5,6 where result should be 1111 1101

unsigned char result = regVal & (0x10 | 0x80);

The way I've attempted to mask above is wrong and I am not sure what operator to use to achieve this.

Have a look at XOR (^ operator) for toggling, at OR (| operator) for setting and at AND (& operator) for clearing. — PhilMasteG, Jan 12 '23 at 11:30
OT: Toggle bits would normally mean change 0 to 1 and change 1 to 0. So 0101 would become 1010 — Support Ukraine, Jan 12 '23 at 11:41
Technically in C you don't have direct access to registers unless you're using inline asm — puppydrum64, Jan 12 '23 at 11:43
Does this answer your question? [How do I set, clear, and toggle a single bit?](https://stackoverflow.com/questions/47981/how-do-i-set-clear-and-toggle-a-single-bit) — justANewb stands with Ukraine, Jan 12 '23 at 11:56
@puppydrum64 Microcontrollers use memory mapped registers. Not every computer is a PC. — Lundin, Jan 12 '23 at 12:19

0___________ · Accepted Answer · 2023-01-12T11:45:52.747

1

To set (to 1) the particular bit:

regVal |= 1 << bitnum;

To reset (to 0) the particular bit:

regVal &= ~(1 << bitnum);

Te write the value to it (it can be zero or one) you need to zero it first and then set

regVal &= ~(1 << bitnum);
regVal |= (val << bitnum);

unsigned char regVal = 0xB5; //1011 0101
// Toggle bits 0-3 (0101) to 1 without changing 4-7 where result should be 1011 1111
regVal |= (1 << 0) | (1 << 1) | (1 << 2) | (1 << 3);

unsigned char regVal = 0x6D; //0110 1101

// Toggle bits 4 and 7 to 1 without changing 1,2,3,5,6 where result should be 1111 1101

regVal |= (1 << 4) | (1 << 7);

unsigned char regVal = 0x6D; //0110 1101

// Set bits 4 to 7 to 1010 without changing 0, 1,2,3, 

regVal &= ~((1 << 4) | (1 << 5) | (1 << 6) | (1 << 7));
regVal |= (0 << 4) | (1 << 5) | (0 << 6) | (1 << 7);

edited Jan 12 '23 at 11:45

answered Jan 12 '23 at 11:33

0___________

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I really wish C would just add binary constants, it would be so much easier to read than this. – puppydrum64 Jan 12 '23 at 11:45
@puppydrum64 gcc has it as an extension: https://godbolt.org/z/nf935TcKs – 0___________ Jan 12 '23 at 11:46
@puppydrum64 They will be added in the upcoming C23. Why, I don't know, since the correct solution is to use named constants for each mask, not magic numbers in any base. – Lundin Jan 12 '23 at 12:14
Honestly, I just find `0b10110011` easier to read than `(1 << 7) | (1 << 5) | (1 << 4) | (1 << 1) | 1` – puppydrum64 Jan 12 '23 at 15:07

score 0 · Answer 2 · answered Jan 12 '23 at 12:17

0

Simply use bitwise XOR. regval ^= 0xFu; Or if you will regval ^= MASK;

And that's it.

answered Jan 12 '23 at 12:17

Lundin

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This is the correct answer for "toggle bits". But OP doesn't really wants to toggle bits. OP wants to "set bits". – Support Ukraine Jan 12 '23 at 13:07
@SupportUkraine Since it says toggle bits at multiple places all over the question, I will assume that they want to toggle bits. – Lundin Jan 12 '23 at 13:35
Kind of agree but OP writes: "Toggle bits 0-3 (0101) **to 1** without changing 4-7 where result should be 1011 **1111**" So OP is using "toggle" incorrectly. OP means "set" instead of "toggle" – Support Ukraine Jan 12 '23 at 13:57
@SupportUkraine Too bad for them... these are industry de facto standard terms. – Lundin Jan 12 '23 at 14:11
Agree. I added that as comment several hours ago – Support Ukraine Jan 12 '23 at 14:13

How can I toggle certain bits in an 8bit register?

2 Answers2