Problems with variable in recursive programming in javascript

Question

I played around with some recursive programming. I have one variable to keep track of my depth (d). The console log is for me to see where the program is at the moment.

var Test = function(){
	this.rekursiv = function(d){
		this.d = d;
		console.log("case 1 granted, d: " + this.d);		
		if( this.d < 3)	{
		console.log("going deeper..");
		this.rekursiv(this.d + 1);
		console.log("going back..");		
		}
		console.log("d: " + this.d  );
		}
}	
t = new Test();
t.rekursiv(0);

Here is my problem: Whenever I go one level deeper, I pass "this.d + 1" to the next level. Howevever, debugging the code (with the console.log) shows that d doesn't only get changed in one level/depth, but in every level/depth.

Why is this so? How can i prevent the code from doing this?

this.d isn't a variable. It's a property on an object, which happens to be the same object used as the value of this in the recursive call. There's no reason for a separate variable in your code, since you already have the d parameter. — user1106925
– user1106925, Commented Sep 4, 2016 at 17:53
You can write it like this: function Test(d) { if (d < 3) { console.log(d); Test(d + 1); console.log(d) } else console.log("base case") }. The problem with your function is, that it doesn't return anything. The base case just performs a side effect (console.log). — user6445533
– user6445533, Commented Sep 4, 2016 at 19:19

Nina Scholz · Accepted Answer · 2016-09-04 19:59:25Z

Why not use the local variable d?

With this.d, you are setting a property of the instance of Test. And with the ending of rekursiv, you do not change the value back to the former value.

var Test = function () {
        this.rekursiv = function(d) {
            console.log("case 1 granted, d: " + d);
            if (d < 3) {
                console.log("going deeper..");
                this.rekursiv(d + 1);
                console.log("going back..");		
            }
            console.log("d: " + d  );
        }
    },
    t = new Test;

t.rekursiv(0);

Another solution would be, to increase this.d at the start of the function rekursivand decrease it at the end.

var Test = function () {
        this.d = 0;
        this.rekursiv = function() {
            this.d++;
            console.log("case 1 granted, d: " + this.d);		
            if (this.d < 3)	{
                console.log("going deeper..");
                this.rekursiv();
                console.log("going back..");		
            }
            console.log("d: " + this.d  );
            this.d--;
        }
    },
    t = new Test;

    t.rekursiv();

gaetanoM · Accepted Answer · 2016-09-04 17:57:29Z

2

Remove this line:

this.d = d;

Now, d is a local variable and no more instance variable.

var Test = function(){
  this.rekursiv = function(d){
    console.log("case 1 granted, d: " + d);
    if( d < 3)	{
      console.log("going deeper..");
      this.rekursiv(d + 1);
      console.log("going back..");
    }
    console.log("d: " + d  );
  }
}
t = new Test();
t.rekursiv(0);

edited Sep 4, 2016 at 17:57

answered Sep 4, 2016 at 17:55

gaetanoM

42.1k6 gold badges45 silver badges63 bronze badges

2 Comments

Michał Młoźniak Over a year ago

Actually declaring d again inside rekursiv function is not necessary. This line can be removed.

gaetanoM Over a year ago

@MichałMłoźniak You are right, it is pointless to redeclare the same variable.

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