CODEWITHSUNDEEP
java
Czechmate
Czechmate

Reputation: 11

Regarding local variables in Java

I have a quick question regarding local variables in Java:

If, upon declaring a local variable, I point it at an instance variable, does the local variable then act as a reference to that instance variable, or does act like a temporary deep copy?

In other words, if I invoke a modifier method on the newly initialized local variable, will the local variable act as a reference and invoke the modifier on the instance variable, will it modify a copied version pointed to by the local variable, or can modifier methods not be invoked on local variables?

Eg.

public static > boolean isSorted(Stack s) {

...(bunch of code)

else if(s instanceof DynamicArrayStack) { ...(bunch of code)

DynamicArrayStack tempStack = (DynamicArrayStack) s; E elem = (E) tempStack.pop();

...(bunch of code) } ...(bunch of code) }

Will invoking pop() on 'tempStack' cause pop() to be invoked on the instance of Stack pointed to by the parameter 's' as well? Or will it just affect the contents of my 'tempStack'?

Upvotes: 1

Views: 192

Answers (6)

calebds
calebds

Reputation: 26228

If, upon declaring a local variable, I point it at an instance variable, does the local variable then act as a reference to that instance variable, or does act like a temporary deep copy?

The first option. Java doesn't implicitly make deep copies of objects. What you have is a temporary reference to the instance object, which the instance variable also references.

Upvotes: 5

Chris Browne
Chris Browne

Reputation: 1612

In your example, "tempStack" and "s" both point to the same instance of DynamicArrayStack (if it is a DynamicArrayStack).

In order for "tempStack" to function as it appears the programmer expects it to, you would need to call some kind of clone/copy method on "s" first.

The Object class (from which all other classes are necessarily extended) provides the "clone" method for this purpose, but you should be warned that it is considered bad practice to use this method because it doesn't provide a deep copy and can introduce some funky bugs as a result; you should use a copy constructor instead.

Upvotes: 2

blitzqwe
blitzqwe

Reputation: 2040

Yes. Remember java works with pointers not values. so when you say

tempStack = s;

tempStack actually gets the value of s, but in java this value is a pointer.

So now both tempStack and s point to the same data.

if you do tempStack.pop() it is as doing s.pop() or vice-versa.

Upvotes: 0

Christian Vielma
Christian Vielma

Reputation: 15995

It will change s and tempStack (because is the same element referenced). But it will not change the object that was passed in isSorted, given that parameters are passed by value in Java.

So if you have:

Stack myStack = new Stack()
...
isSorted(myStack);

The object myStack will not be modified. But if in your method isSorted s and tempStack will.

I mean:

s.pop()

will be the same than

 tempStack.pop()

Both variables reference the same object so both will modify the same object.

This would clarify a little bit more.

Best regards,

Upvotes: 0

jabu.10245
jabu.10245

Reputation: 1892

Short answer: yes, s and tempStack are two variables referencing the same instance.

Calling tempStack.pop() is the same as calling s.pop(), since both reference the same object.

Upvotes: 0

Jens Schauder
Jens Schauder

Reputation: 81907

Its all references. So the pop you mentioned will affect s as well since it is the same object.

Upvotes: 0

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