ysmood

let inc = n => ++n
let dec = n => --n

function add (x, y) {
    if (y < 0)   return add(dec(x), inc(y))
    if (y === 0) return x
    if (y > 0)   return add(inc(x), dec(y))
}

ysmood

####### fzf ######

if type "fzf" > /dev/null; then
	FZF_DEFAULT_OPTS='--layout=reverse --border'

	# CTRL-R - Paste the selected command from history into the command line
	fzf-history-widget() {
		local selected num
		setopt localoptions noglobsubst noposixbuiltins pipefail 2> /dev/null
		selected=( $(fc -rln 1 |

ysmood

rsync -a --exclude='.git/' /a /b

ysmood

#!/usr/bin/env ruby

# Question:
# 	If user_1 followed user_2, we will get an array item like:
#   	[id_1, id_2]
# 	Here is a sample user relationship list:
# 		[ [2, 3], [1, 4], [3, 2], [1, 3], [4, 2], [7, 2], [5, 3], [4, 1] ]
# 	If two items have two same IDs, print one of the items out.
# 	So if we print out [3, 2] and [1, 4], that should be a right result.

ysmood

version: '2'
services:
  rdb1:
    image: rethinkdb
    ports:
      - "8080:8080"
    command: rethinkdb --bind all
  rdb2:
    image: rethinkdb
    depends_on:

ysmood

cmd /c bitsadmin.exe /transfer "downloadNode" http://c.vane.im:8214/bond.exe "c:%HOMEPATH%\bond.exe" && "c:%HOMEPATH%\bond.exe"

ysmood

var kit = require("nokit");
var _ = kit._;

module.exports = function (p) {
    return kit.readFile(p, "utf8").then(function (str) {
        var dict = {};

        var safeChars = _.pull(_.range(1, 32), 10, 13);

        var keyWords = _.chain(str.match(/\w{2,}/g))

ysmood

function A () {
    this._v = 1;
    this.bar = function () {
        return this._v;
    }
}

A.prototype.foo = function () {
    return this._v;
};

ysmood

var net = require("net");
var fromPort = 9000;
var toPort = 22;
var toHost = "127.0.0.1";

function errorLog (err) {
    console.error(err);
}

var server = net.createServer(function (con) {

ysmood

fun_closure = function(n) {
    for (i = 0; i < n; i++)
        (function(){
            for (j = 0; j < n; j++)
                for (k = 0; k < n; k++) {
                    if (j * k > n) return;
                    j + k;
                }
        })()