Gray code
In ordinary binary, counting from 7 (0111) to 8 (1000) flips four bits at once. If those bits are read by a mechanical encoder or sampled by a clock while they change, any mix of old and new bits can appear, and a momentary 1111 (15) is a large error. Gray code (reflected binary code) orders the numbers so that consecutive values differ in exactly one bit. Rotary encoders, Karnaugh maps, some ADCs and error-correction schemes use it.
The conversion is one line each way:
gray = n ^ (n >> 1)
binary = g; while (g >>= 1) binary ^= g; // XOR with all right shiftsBCD
Binary-coded decimal stores each decimal digit in its own 4-bit nibble: 1234 becomes 0001 0010 0011 0100, which is 0x1234 as a hex number. It wastes some space (nibbles A–F are never used) but makes decimal display and exact decimal arithmetic trivial, which is why clocks, calculators, financial COBOL and the real-time clock chip in every PC use it. Reading a BCD byte as plain binary gives nonsense, which is the usual bug.
Popcount and parity
The Hamming weight (popcount) is the number of 1 bits. It is a single instruction on modern CPUs (POPCNT) and the basis of Hamming distance, bitset cardinality and some hashing tricks. Parity is popcount mod 2; a parity bit is the simplest error-detecting code, catching any single flipped bit.
// Kernighan's trick: each step clears the lowest set bit
int count = 0; while (n) { n &= n - 1; count++; }Questions
Why do rotary encoders use Gray code?
Between two adjacent positions only one contact changes, so a reading taken mid-transition is always one of the two neighbouring values, never a wild one. Binary encoders can show any value during a multi-bit transition.
Is Gray code unique?
The reflected binary code is the standard one, but any ordering where neighbours differ in one bit is “a” Gray code. Balanced and n-ary Gray codes exist for special uses.
How do I convert BCD to binary in code?
Take each nibble as a digit: value = value*10 + nibble, from the most significant nibble down. Or, for two-digit bytes, ((b >> 4) * 10) + (b & 0x0F). Many RTC chips need exactly this.
What is the fastest way to count bits?
Use the built-in: __builtin_popcount (C), bits.OnesCount (Go), int.bit_count() (Python 3.10+), u32::count_ones (Rust). They compile to a single instruction where available.