I'm looking at the as3delaunay library and most of the code is clear to me. This part is not, however (note the line that I put preceded with an arrow):
public function circles():Vector.<Circle>
{
var circles:Vector.<Circle> = new Vector.<Circle>();
for each (var site:Site in _sites)
{
var radius:Number = 0;
var nearestEdge:Edge = site.nearestEdge();
=======>> !nearestEdge.isPartOfConvexHull() && (radius = nearestEdge.sitesDistance() * 0.5);
circles.push(new Circle(site.x, site.y, radius));
}
return circles;
}
For reference, isPartOfConvexHull() is found in Edge.as and looks like this:
internal function isPartOfConvexHull():Boolean
{
return (_leftVertex == null || _rightVertex == null);
}
What does !nearestEdge.isPartOfConvexHull() do? Does that mean that the radius = nearestEdge.sitesDistance() * 0.5 only executes if false is returned from the call to isPartOfConvexHull()? Does that stop execution of any other code?
It is equivalent to:
if (!nearestEdge.isPartOfConvexHull()) {
radius = nearestEdge.sitesDistance() * 0.5;
}
In the following line:
var b:Boolean = expression1 && expression2;
expression2 will not be evaluated if expression1 is false because we already know the final result: b = false.
Now in the following line:
expression1 && expression2;
The same thing happens except the fact that we are not assigning the result to a variable.
And this is exactly what happens in the line you are asking about where !nearestEdge.isPartOfConvexHull() is the first expression and (radius = nearestEdge.sitesDistance() * 0.5) is the second expression.
To extends #sch answer with some explanations (I didn't knew if editing answer to almost double it was ok).
This is based on lazy execution of the interpreter. If (!nearestEdge.isPartOfConvexHull()) is False then there's no need to execute the second part of the AND statement to know it'll be False, then it's left unexecuted. If it's true the evaluation of the complete statement is needed (and then done) to tell wether or not this boolean is True. So this is equivalent to an if statement.
TMHO this is bad code since it's to much condensed and hard to understand.
Related
I have an array of Nodes 'flags', and I want to set my object's position at the first object in that array, it works and the object actually gets positioned as intended, but when I make the comparison it fails and logs 'NO'.
The line of code that sets the position works, but the comparison fails, what's wrong here?!
start: function () {
this.node.position = this.flags[0].position;
this.movement();
},
movement: function() {
if (this.node.position == this.flags[0].position) { // Problem
console.log("YES");
}
else {
console.log("No");
Update:
When I do it like this it works:
if (this.node.position.x == this.flags[0].position.x) // or position.y
Well if you write javascript here (and it looks like you do) there're two things you should know:
You can't compare objects with == out of the box
({"a":1} == {"a":1})
Will return false (you may try it yourself in your browser.
As a workaround you could do something like:
function posCompare(p1, p2){
return p1.x === p2.x && p1.y === p2.y;
}
Then use it instead of == for positions
See how I use === instead of ==? Second thing to know is Use only ===. You can learn the difference Which equals operator (== vs ===) should be used in JavaScript comparisons? but I'd keep away from == anywhere. It's slower, it may cause strange errors here and there - just don't use it at all
I am working on a standalone interface.
I have several text boxes, and all have Names and IDs.
If I run this code:
var re = /someregexp/g;
var k ="sometext";
textBoxOne.setText(re.test(k)); //textBoxOne was selected first
The correct result 'true' or 'false' is displayed.
asking for a 'typeof(re.test(k))' correctly returns boolean.
However this code:
if (re.test(k)) {
textBoxTwo.setText("matched.");
} else {
textBoxTwo.setText("NOT matched.");
}
Always goes into the 'else' branch. if (re.test(k) == true) renders the same result.
Looks like a bug to me, anyone else also found this?
Edit: AdamL has pointed out (thanks Adam), in his comments below, that there is more to this than I supposed. I leave my original code below for reference.
The myFunction2() below follows the "matched" branch as you would expect a correctly working regex (so perhaps this can be used to work around the bug):
function myFunction2() {
var k ="sometext";
Logger.log( /ome/g.test(k) );
if ( /ome/g.test(k) ) {
Logger.log("matched.");
} else {
Logger.log("NOT matched.");
}
}
Conversely, as you have observed, the myFunction() below unexpectedly follows through to the "NOT matched" branch.
function myFunction() {
var re = /ome/g;
var k ="sometext";
Logger.log(re.test(k));
if (re.test(k)) {
Logger.log("matched.");
} else {
Logger.log("NOT matched.");
}
}
I'm working on a match-3 style puzzle game using Flixel, and so I'm working on checking each row and column to see if there is a match at any given time. However, I have 6 different pieces (as of right now) that are active, and each piece is identified by an integer. Given that, I can check, for each and every single piece, by doing something like this:
public function matchingCheck():void
{
if (piecesArray[0][1] == 1 && piecesArray[1][1] == 1 && piecesArray[2][1] == 1) {
FlxG.log("Yay!");
}
}
However, this is rather unwieldy and would basically cause way too much repetition for my liking.
At the very least, I would like to be able to check if the values in these arrays are equal to one another, without having to specify which value it is. At the very best, I'd love to be able to check an entire row for three (or more) adjacent pieces, but I will settle for doing that part manually.
Thanks for your help!
EDIT: Nevermind, my edit didn't work. It was just checking if piecesArray[2][1] == 1, which makes me a sad panda.
EDIT 2: I've selected the correct answer below - it's not exactly what I used, but it definitely got me started. Thanks Apocalyptic0n3!
You could cut down on that code a little bit by using another function
private function checkValid( arrayOfItemsToCheck:Array, value:* ):Boolean {
for ( var i:Number = 0; i < arrayOfItemsToCheck.length; i++ ) {
if ( arrayOfItemsToCheck[i] != value ) {
return false;
}
}
return true;
}
Then you just do this in your if statement:
if ( checkValid( [ piecesArray[0][1], piecesArray[1][1], piecesArray[2][1] ], 1 ) ) {
FlxG.log("Yay!");
}
That does assume all items need to be equal to 1, though. It's still a lot of code, but it cuts out one set of "= 1 &&" for each check.
How about something like this which would tell you both if a match existed and what match it was:
public function checkForMatch():void{
var rows:int = piecesArray.length;
for(var i:int=0; i<rows; i++){
var match:int = checkRow(piecesArray[i]);
if(match > -1) {
FlxG.log("Yay you matched " + match);
}
}
}
private function ckeckRow(row:Array):int{
if(row[0] == row[1] == row[2]){
return row[0];
}
return -1;
}
In this example I'm trying to create an Array of length 5 where each ellement contains the number of times .3 can be summed without exceeding 1. i.e. 3 times. So each element should contain the number 3. Here is my code:
Array[(
workingCount = 0;
workingSum = 0;
done = false;
While[! done,
workingSum = workingSum + .3;
If[workingSum > 1, done = true; workingCount, workingCount++]
])
, 5]
In the 3rd to last line there I have workingCount without a ; after it because it seems like in Mathematica omitting the ; causes the value a statement resolves to to be returned.
Instead I get this:
{Null[1], Null[2], Null[3], Null[4], Null[5]}
Why does this happen? How can I get my program to do what I want it to do? i.e. In the context of the function passed to Array to initialize it's elements, how to I use complicated multi-line functions?
Thanks in advance.
Two things:
First, one way to be able to do that in Mathematica is
Array[
Catch[
workingCount = 0;
workingSum = 0;
done = False;
While[! done,
workingSum = workingSum + .3;
If[workingSum > 1,
done = True; Throw#workingCount,
workingCount++]]] &,
5]
Second, and most important: you never should do that in Mathematica! Really.
Please visit for example the Stack Exchange site for Mathematica, and read the questions an answers there to get some grip on the programming style.
Your problem comes from the fact that you are trying to initialize your array, but are trying to do so without an explicit function call - which is what you need to do.
See here for documentation on Arrays in Mathematica:
http://reference.wolfram.com/mathematica/ref/Array.html
That aside, and minor errors (True and False have to be capitalized), this is what you want to do:
f[x_] :=
(
workingCount = 0;
workingSum = 0;
done = False;
While[done != True, workingSum = workingSum + 0.3;
If[workingSum > 1, done = True, workingCount++]
];
Return[workingCount];
);
Array[f, 5] (* The array here is generating 5 values of the return value of f[x_] *)
Here's a example that I've to use when I want to create a button with mouse-over effect:
this.buttonExample.buttonMode = true;
this.buttonExample.useHandCursor = true;
this.buttonExample.addEventListener(MouseEvent.CLICK,myaction);
I'm new to AS3 - is there any way, to simplify this code like this:
this.buttonExample.buttonMode = true;.useHandCursor = true;.addEventListener(MouseEvent.CLICK,myaction);
why does it not works ?
Its already as simple as it gets. Firstly
this.buttonExample.buttonMode = true;
this.buttonExample.useHandCursor = true;
this.buttonExample.addEventListener(MouseEvent.CLICK,myaction)
is much more readable than
this.buttonExample.buttonMode = true;.useHandCursor = true;.addEventListener(MouseEvent.CLICK,myaction);
Always go for readbility over anything else. And secondly,
this.buttonExample.buttonMode = true;
does not return an object so you can't interact with anything.
If you're using that pattern a lot, you can make a helper function:
public function setAsButton(button:Sprite, clickHandler:Function):void {
button.buttonMode = button.userHandCursor = true;
button.addEventListener(MouseEvent.CLICK, clickHandler);
}
Then call it somewhere:
setAsButton(this.buttonExample, myaction);
If you feel that typing this.buttonExample over and over again is too repetitive, simply assign that object to a variable and use that variable in the rest of the statements:
var b : Button = this.buttonExample;
b.buttonMode = true;
b.useHandCursor = true;
b.addEventListener(...);
As other's have mentioned, there's also the with statement, but it's use is discouraged since it makes the code harder to read, and may lead to weird results:
with (this.buttonExample) {
buttonMode = true;
useHandCursor = true;
addEventListener(...);
}
You can, of course, combine these suggestions with other tricks, like chaining assignments:
var b : Button = this.buttonExample;
b.buttonMode = b.useHandCursor = true;
b.addEventListener(...);
Be very careful to only chain assignments in this way if the assigned value is immutable (e.g. true, false, numbers and strings, but not arrays or most other objects), because the same object will be assigned to all variables on the left side. If the value is immutable this doesn't matter, but if it's mutable you can end up with weird results, like this in this example:
a = b = [ ];
a.push(1);
b.push(2);
trace(a); // outputs 1, 2
trace(b); // also outputs 1, 2
The reason for this result is that a and b both reference the same array, and since arrays are mutable it doesn't matter how you access the object, it will still be changed. a and b don't reference different arrays just because they are different variables.
You may think that you could do something like the following, but it will not work.
// this will NOT work
var b : Button = this.buttonExample;
(b.buttonMode = b.useHandCursor = true).addEventListener(...);
The reason why it works to say b.buttonMode = b.useHandCursor = true, but not to add .addEventListener(...) is that the value of an assignment expression (e.g. b.buttonMode = true) is the value assigned to the left hand side (e.g. true). If you add .addEventListener(...) to that you are essentially saying true.addEventListener(...), which clearly is not what you want. In other words
b.buttonMode = b.useHandCursor = false;
is equivalent to
b.useHandCursor = false;
b.buttonMode = b.useHandCursor;
Which should hopefully also make the caveats mentioned above plain.
you can use the with statement. however I'd not encourage you to do so, since it leads to a lot of ambiguity and unclearness.
also, you can have multiple assignments:
this.buttonExample.buttonMode = this.buttonExample.useHandCursor = true;
this sometimes is useful, but for the sake of readability, you shouldn't overuse it.
greetz
back2dos