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

最新推荐文章于 2021-02-08 09:28:45 发布

原创最新推荐文章于 2021-02-08 09:28:45 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#优雅代码

Quora问答专栏收录该内容

1 篇文章

订阅专栏

本文汇集了多位专家分享的经典算法及编程技巧，包括位操作、欧几里得算法、二分查找、快速逆平方根等，涵盖了高效计算和数据处理的方法。

@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++);

（揸枪注：这是复制字符串的代码，在《C程序设计语言》那本书里。）
@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: