md2perpe

Got an idea of a new assignment operator ?= in PHP.

Instead of writing

if (!$x)
    $x = $y;

one could use ?= and write

$x ?= $y;

md2perpe

md2perpe

class Rectangle
{
    protected double width, height;

    constructor(double width, double height)
    {
        this.width = width;
        this.height = height;
    }

md2perpe

<?php

class Base
{
  private $x;

  public function getX()
  {
    return $this->x;
  }

md2perpe

<?php
$tweet = 'Detta är ett #test för min blog... #ignoreme #lördag #måla #mäktigt #coolt #åsa-nisse #()lol()';

$pattern = '/#(\pL+)/';

$replacement = '<a href="#$1">$0</a>';

echo preg_replace($pattern, $replacement, $tweet);

md2perpe

var M = function() { };
M.prototype = {
    m : function() { return 42 }
};

var inst = new M();
inst.f = function() { return 42 };

md2perpe

// Recursive version

function isMeasurable(target, weights)
{
    if (weights.length == 0)
        return target == 0;

    var first = weights.shift();
    return 
        isMeasurable(target-first, weights) ||

md2perpe

Att stödja en regeringen är inte samma sak som att sitta i regeringen. Flyborg påstod att MP var "i regeringsställning", vilket betyder att man verkligen sitter i regeringen.

md2perpe

data Tree t = EmptyTree
            | Node t (Tree t) (Tree t)
			deriving Show

nivel :: Int -> Int
nivel n = nivel' n 0
where
	nivel' 0 _ = 0
	nivel' 1 x = x
	nivel' n x = nivel' (n `div` 2) (x + 1)

md2perpe

firstNonRep :: String -> Maybe Char
firstNonRep cs = fmap fst $ find (\(c, n) -> n == 1) $ foldl count [] cs
	where 
		count []               c             = [(c, 1)]
		count (p@(c', n) : ps) c | c == c'   = (c', n+1) : ps
		                         | otherwise = p : count ps c