setting by reference

@oaishmSep 20.2011

Is there a way to do this php equivalent in javascript:

a=5;
b=&a;

b++;

alert(a) //6
alert(b) //6

I’d like to traverse a binary tree in javascript which isn’t really fond of binary trees.

to post a comment

@Jeff_MottSep 20.2011 — #Not directly. You'll have to wrap the value in an object, because objects are passed by reference.

<i>
 </i>var a = {
 value: 5
 };
 
 var b = a;
 
 b.value++;
 
 alert(a.value); // 6
 alert(b.value); // 6

@oaishmauthorSep 20.2011 — #Yes, I can see how this works in the trivial case. How if I want to traverse a tree

Something like

[CODE]var x={}
 for (i=1;i<100;i++){
 var g=g||x;
 g[i]={};
 g=&g[i]; // by reference to get one level down this is what I really want
 
 }
 [/CODE]

so by the end we'd have x={1:{2:{3:{4: etc.

So we can traverse wtih

[CODE]
 var p = function (n){
 r = n||x;
 if(functiontotestifobject(r)){
 for(var i in r){
 var z = p(r[i]);  //I don't know, something like this, obviously I concocted this
 } else { return something}
 }
 [/CODE]

so that var l = p(); would make l equal something

@Jeff_MottSep 20.2011 — #Yes, I can see how this works in the trivial case. How if I want to traverse a tree

Something like

[CODE]var x={}
 for (i=1;i<100;i++){
 var g=g||x;
 g[i]={};
 g=[color=red]&[/color]g[i]; // by reference to get one level down this is what I really want
 
 }
 [/CODE]

so by the end we'd have x={1:{2:{3:{4: etc.[/QUOTE]

Just remove the ampersand, and you'll have it.

@rnd_meSep 20.2011 — #

[CODE]a=[5];
 b=a;
 b[0]++;
 alert(a);[/CODE]

@martingalehSep 23.2011 — #we can have

a={1:a, 2:b} whatever

b=a

b=b[2]

c={a:4}

b=c

alert(a[2]['a']) //woudlnt' this show undefined because we would have liked to do

b=&b[2] to differentiate between when we set a pointer to point to something else or when we point to something else.

setting by reference

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