XOR for toggling bits in Embedded C - Step-by-Step Execution

Concept Flow - XOR for toggling bits

Start with initial bits

↓

Apply XOR with toggle mask

↓

Bits where mask=1 flip

↓

Bits where mask=0 stay same

↓

Result: toggled bits

↓

End

XOR flips bits where the mask has 1s and leaves others unchanged, toggling bits step-by-step.

Execution Sample

Embedded C

unsigned char bits = 0b10101010;
unsigned char mask = 0b00001111;
bits = bits ^ mask;

This code toggles the lower 4 bits of 'bits' using XOR with 'mask'.

Execution Table

Step	bits (binary)	mask (binary)	XOR Result (binary)	Explanation
1	10101010	00001111	10100101	Initial bits XOR mask toggles lower 4 bits
2	10100101	00001111	10101010	Applying XOR again toggles lower 4 bits back
Exit	-	-	-	No more toggling, bits restored to original

💡 After two toggles, bits return to original value, showing XOR toggling effect

Variable Tracker

Variable	Start	After 1	After 2	Final
bits	10101010	10100101	10101010	10101010
mask	00001111	00001111	00001111	00001111

Key Moments - 2 Insights

Why does applying XOR twice with the same mask restore the original bits?

Because XOR flips bits where mask=1, doing it twice flips them back, as shown in execution_table rows 1 and 2.

What happens to bits where the mask has 0?

Those bits stay the same since XOR with 0 leaves bits unchanged, as seen in the unchanged upper 4 bits in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'bits' after the first XOR operation?

A 10101010

B 10100101

C 00001111

D 11111111

💡 Hint

Check execution_table row 1 under 'XOR Result (binary)'

At which step does 'bits' return to its original value?

A Step 1

B Exit

C Step 2

D Never

💡 Hint

Look at execution_table rows 1 and 2 for 'bits' values

If the mask was 0b11110000 instead, which bits would toggle?

A Upper 4 bits

B All bits

C Lower 4 bits

D No bits

💡 Hint

Mask bits set to 1 determine which bits toggle; see variable_tracker for mask bits

Concept Snapshot

XOR toggles bits where mask bits are 1.
Applying XOR twice with same mask restores original bits.
Bits with mask=0 stay unchanged.
Use: bits = bits ^ mask;
Common in embedded C for bit toggling.

Full Transcript

This visual execution shows how XOR toggles bits in embedded C. Starting with bits 0b10101010 and mask 0b00001111, XOR flips the lower 4 bits. After the first XOR, bits become 0b10100101. Applying XOR again with the same mask flips those bits back to 0b10101010. Bits where mask is 0 remain unchanged throughout. This demonstrates XOR's property of toggling bits and restoring original values when applied twice with the same mask.