preslavrachev

/* we are using the assumption that every entity is constructed recursively as every child component is given an id of +1

 so it could be that my id is #5, but if I have 7 child components, the ID of my next neighbor (the next entity constructed by my parent, which is on the same tree level ) will be #13, because I will be constructed recursively to the lowest level first */


public static void sendMessage(String methodName, Object args)
        {

            int id = 5;
            int parentId = 0;

preslavrachev

var c = "it ws a #nice and #sunny day in Berlin";
c.match(/\B#\w*[a-zA-Z]\w*/g); //should return #nice and #sunny

preslavrachev

var p = function() {
	console.log("Executing the callback function");
}

var c = function(callBack) {
	console.log("Executing the wrapper function");
} 

c(p);

preslavrachev

#include <iostream>
using namespace std;

void theWrapperFunction(int (*fnc) ()) {
	cout << fnc();
}

int theCallBackFunction() {
	return 12345;
}

preslavrachev

#include <iostream>
#include "pixel.h"

using namespace std;

typedef int (*FunctionFunc)(); // check that out!!!

void theWrapperFunction(FunctionFunc fnc) {
	cout << fnc();
}

preslavrachev

class TestClass {
	var a:String;
	
	public function new() {
		this.p = "test1";
	}
}

....

preslavrachev

class TestClass implements Dynamic { // this is the crucial part
	var a:String;
	
	public function new() {
		this.p = "test1";
	}
}

....

preslavrachev

#in the Extension folder:

haxelib run hxcpp Build.xml -Diphoneos #device

#or

haxelib run hxcpp Build.xml -Diphonesim #simulator

#or

preslavrachev

var x = 2 // x gets a reference to the value of the scalar 2;

var y  = x  // also gets a reference to the value of the scalar 2; Please, note that y does not refer to x, but //to the value to which x refers. That will make more sense in the later examples.

x++;

y = ??? x = ??? // x == 3, while y==2, because x was given a new reference to the value of the scalar 3. Since //y points a reference to the value of the scalar 2, and not to x, it keeps the value

preslavrachev

SELECT uid,name,is_app_user FROM user WHERE is_app_user AND uid IN (SELECT uid2 FROM friend WHERE uid1 = me())