You can also use:
combo=0;
if(a){combo+=1000;}
if(b){combo+=100;}
if(c){combo+=10;}
if(d){combo+=1;}
Then convert combo to a string and check the character at each
position ( charAt(0), charAt(1), etc.) for a one or zero, which will
tell you if the associated variable is true or false.
-Marc
At 10:47 AM 1/25/2006, you wrote:
You seem to have something like a 'Round Robin' situation, only more complex
as it is not only about pairs, but also about singles/triplets/quadruplets.
If each of these possible combinations leads to a different action, the
if/else if or switch statement will naturally be very long...
To make it quite a bit more readable and thus managable, maybe you could do
combo = '';
if ( a ){ combo += 'a'; }
if ( b ){ combo += 'b'; }
if ( c ){ combo += 'c'; }
if ( d ){ combo += 'd'; }
switch(combo){
case 'a':
// do stuff
brek;
case 'ab':
// do some other stuff
break;
case 'abc':
//etc.
}
For finding all the 'pairs' see my RoundRobin class below (used to be in the
archives, but can't find it anymore)
hth
--------------
Andreas Weber
motiondraw.com
class com.motiondraw.util.RoundRobin{
/*
Usage
import com.motiondraw.util.RoundRobin
numPlayers = 12;
// create players
players = new Array();
for(var i=0; i<numPlayers; i++){
players[i] = String.fromCharCode(65+i);
}
var rr:RoundRobin = new RoundRobin(players);
rr.displaySchedule();
*/
public var players:Array;
public var schedule:Array;
function RoundRobin(players:Array){
init.apply(this, arguments);
}
function init(players){
this.players = players;
var pc = players.slice(0);
var n = pc.length;
if(n%2){
var odd = true;
pc.push('@');
n++;
}
var nr = n - 1;
var r = new Array();
schedule = new Array();
for (var i=0; i<nr; i++) {
r[i] = new Array();
r[i][0] = pc[nr];
for (var j=0; j<n-1; j++) {
r[i][j+1] = pc[(i+j) % (n-1)];
}
}
for(var i=0, len=r.length; i<len; i++){
schedule[i] = new Array();
for(var j=0; j<nr/2; j++){
if(!(odd && j==0)){
schedule[i].push([r[i][j],
r[i][nr-j]]);
}
}
}
}
public function displaySchedule(short:Boolean){
var output = new Array();
for(var i=0, len=schedule.length; i<len; i++){
//trace('--- '+(i+1)+'. round ---');
for(var j=0, len2=schedule[i].length; j<len2; j++){
//trace(' '+j+'
'+schedule[i][j][0]+' vs. '+schedule[i][j][1])
output.push('"'+schedule[i][j][0]+''+schedule[i][j][1]+'"');
}
}
trace(output);
trace(output.length);
}
}
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of eric
dolecki
Sent: Wednesday, January 25, 2006 7:03 PM
To: Flashcoders mailing list
Subject: [Flashcoders] checking combinations
I have 4 variables I need to check the states of routinely... and
combinations thereof.
I am assuming that if I have a counter, and interogate and += them values, I
can then check the value of the counter to determine the combinations.
psudeo-code:
var counter:Number = 0;
if ( a ){ counter += 2; }
if ( b ){ counter += 3; }
if ( c ){ counter += 6; }
if ( d ){ counter += 12;}
if ( counter == 2 ){
// only a was true
} else if ( counter == 3 ){
// only b was true
} ...
Which is fine, but thats gonna be one honkin' if else statement to catch all
the combinations.
Is there a better way of doing this that will catch all the possible
combinations in an elegant way?
- edolecki
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