this keyword

[Amie Mac]

I have the below discussion question posted for my class. I am pretty sure I understand the use of the this keyword. I stated that when using this, it is referencing that particular method. What I don't understand is the 5 instantiated objects and 5 different variables. Can someone help explain? Or provide an example of what the below statement is showing. I can't picture what is being asked here.

An instance method might contain a statement like this.weight = 1.0; but if that method's class currently has five instantiated objects, there are five different variables called weight. How can we determine which one is getting the new value?


Thanks
Amie

[Knute Snortum]

The keyword this refers to the current object, not a method. An instantiated object just means you have used to the new keyword to create an object. Does this help clarify the question?

[Campbell Ritchie]

Each of the five objects has a weight. They do not necessarily weigh all the same. You use this.weight to distinguish the weight of the current object from any other weight.

If you have not declared a local variable called weight or a parameter called weight, then the only possible weight you can find turns out to be this.weight. In which case writing this. is redundant. But some people like to write this. regardless so as to make absolutely sure they are referring to the weight of the current object (i.e. the instance field of that name). If you write this.fieldName you always get the field as opposed to local variable/parameter.

[Amie Mac]

So if I understand you correctly there are 5 different object all with a weight variable?

Would it be like this:

Public weightObject1 (double weight) 
{
    this.weight = 1.0;
}

Public weightObject2 (double weight) 
{
    this.weight = 2.0;
}

Public weightObject3 (double weight) 
{
    this.weight = 10.0;
}

etc?

Or how do you make 5 instances of an object?

[Campbell Ritchie]

No.

WeightObject w1 = new WeightObject(1.23);
WeightObject w2 = new WeightObject(2.34);
WeightObject w3 = new WeightObject(3.45);
WeightObject w4 = new WeightObject(4.56);
WeightObject w5 = new WeightObject(5.67);

Obviously you won't want those 5 variables lying loose like that; you will want to put them into a data structure.

[Campbell Ritchie]

Amie Mac wrote:So if I understand you correctly there are 5 different object all with a weight variable? . . .

Yes.

[Campbell Ritchie]

You don't make instances of an object. You make instances of a class.

[Amie Mac]

Campbell Ritchie wrote:No.

WeightObject w1 = new WeightObject(1.23);
WeightObject w2 = new WeightObject(2.34);
WeightObject w3 = new WeightObject(3.45);
WeightObject w4 = new WeightObject(4.56);
WeightObject w5 = new WeightObject(5.67);

Obviously you won't want those 5 variables lying loose like that; you will want to put them into a data structure.

So these are instances of the class WeightObject correct?
If you had a weight variable declared like:

double weight;

could the instance then look like this:

WeightObject w1 = new WeightObject(weight);
    this.weight=1.0;

?

I am trying to understand it as the discussion question is written.

[Campbell Ritchie]

Amie Mac wrote: . . .
could the instance then look like this:
. . ..

No.

You never see what an instance “looks like”; you only ever see the class it was created from.

class WeightObject
{
   private double weight;

   public WeightObject(double weight)
   {
      this.weight = weight;
   }

   // usual range of methods
}