[Quora] What is the most elegant line of code you've seen?

@Murtaza Aliakbar:

// Counting # bits set in 'v', Brian Kernighan's way
for (c = 0; v; c++)    v &= v - 1;

@Joshua Seims:
Euclid's Greatest Common Divisor algorithm (considered the oldest historical algorithm):

function (a, b) { return b ? gcb(b, a % b) : a; }

@Pradeep George Mathias:
Binary Search (The main "elegant"point is in line 3.)

int lo = 0, hi = N;
for (int mid = (lo + hi) / 2; hi - lo > 1; mid = (lo + hi) / 2)
    (arr[mid] > x ? hi : lo) = mid;
return lo;
// assuming N < 2^30, else "(lo + hi)" may overflow int
// can be worked around using lo + ( hi - lo) / 2

@Ravi Tandon:
Checking whether a (natural number > 1) is a power of two or not

return (!(x & (x - 1)));

@Vinay Bajaj
Swaps A and B:

A = A + B - (B = A);

@Jesse Tov:

while (*s++ = *t++);

(銃注:これは文字列をコピーするコードです.『Cプログラム設計言語』の本にあります.)
@Sugavanesh Balasubramanian:
Not exactly a line of code, but this is a terminal command 
(*Disclaimer: Don't ever try this command! - Reference: :(){ :|:& };:)

:(){ :|:& };:

@Chetan bademi:
Inverse square roots are used to compute angles of incidence and reflection for lighting and shading in computer graphics. This code uses integer operations, avoiding computationally expensive floating point operations, which makes it four times faster than normal floating point division. 

float FastInvSqrt(float x) {
    float xhalf = 0.5f * x;
    int i = *(int*) &x;                 // evil floating point bit level hacking
    i = 0x5f3759df - (i >> 1);          // what the fuck?
    x = *(float*) &i;
    x = x * (1.5f - (xhalf * x * x));
    return x;
}

@Thomas Le Feuvre:
One line sudoku solver(python)

def r(a):i=a.find('0');~i or exit(a);[m in[(i-j)%9*(i/9^j/9)*(i/27^j/27|i%9/3^j%9/3)or a[j]for j in range(81)]or r(a[:i]+m+a[i+1:])for m in'%d'%5**18]from sys import*;r(argv[1])

This is a 
recursive algorithm for solving a sudoku. I don't think it's incredibly efficient though. It's a fun 
one to work out what it's doing. If you are wanting to see a deep explanation of it see here:http://stackoverflow.com/questions/201461/shortest-sudoku-solver-in-python-how-does-it-work
@Brien Colwell:
For a 64-bit integer x,

for m in [0xff51afd7ed558ccdL, 0xc4ceb9fe1a85ec53L, 1]: x = (x ^ (x >>> 33)) * m

Applies t
h
e M
urmurHash3 b
it avalan
che to generate a near-uniform distribution for x drawn from nearly any input distribution. Useful for uniformly distributing processing and data, much more efficient than md5 and other deep hashes.
@Jiten Thakkar:

while(x-->0) { /* Do something */ }

This gives a feeling of x tends to 0.
しばらくはこれだけ抜粋しましょう.これらは私が相対的に理解できる言語で書いたコードで、原文にはperlとHaskellが書いたコードもあります.私は分からないので、貼っていません.質問全体の答えに興味のある方はこちらを訪問できます.