CODEWITHSUNDEEP
javapointers
NewbieDave
NewbieDave

Reputation: 1259

How to implement a basic pointer

I know here is no pointer in Java. But how do I change a value in the calling scope? For instance, I want to write a function that takes an integer num, set the integer to 0 if it's greater than 21, otherwise do nothing. In the main, my code is as follow:

int a=34;
KillOver21(a);
System.out.print(a);

I expect an 0.

Upvotes: 3

Views: 148

Answers (4)

Subhrajyoti Majumder
Subhrajyoti Majumder

Reputation: 41200

It is simply not possible, Java supports pass by value. int a's value will be copied to the function.

You could use Object instead of primitive where the reference value will be copied to your function by which you can get the actual object and modify it.

Upvotes: 3

bsorrentino
bsorrentino

Reputation: 1641

The simpliest way (quick&dirty) is to put value within an array

int holder[] = new int[]{ a};

KillOver21(holder)

System.out.printf( "value=[%d]", holder[0]  );

void KillOver21(int holder[] ) {
    holder[0] = 0;
 }

Upvotes: 1

sanbhat
sanbhat

Reputation: 17622

Java is pass by value, so a copy of the parameter a is sent to the method, so modification to a in the method will not affect the original argument a in main

The max you can do is return int from KillOver21(a) method

int z = KillOver21(a); // This will return 0
System.out.print(z);

But you can achieve something like that with custom objects, say you have a class

class AHolder {

    public int a;

}

then you can expect AHolder instance to change

public static void main(String [] args) {
     AHolder a = new AHolder();
     a.a = 34;
     killOver21(a);
     System.out.println(a.a);
}

public static void killOver21(AHolder b) {
    if(b.a > 21) {
       b.a = 0;
    }
}

Since in the latter (even if its Pass by Value) , the reference is copied and both reference point to same object. So changes made inside the killOver21 method actually changes the object.

enter image description here

Upvotes: 4

jason
jason

Reputation: 241641

Fundamentally impossible in Java, period. int are immutable, and passed by value. You would need to create a mutable int type:

class MutableInt {
    private int value;

    public MutableInt(int value) { this.value = value; }

    public getValue() { return this.value; }
    public setValue(int value) { this.value = value; }
}

Then:

void KillOver21(MutableInt m) {
    if(m.getValue() > 21) { m.setValue(0); }
}

However, be aware the mutable types that represent concepts that are defined by their value rather than their identity are generally an extremely bad idea. But, this is the only way to achieve what you're trying to achieve. Again, I caution you with the strongest words: what you're doing is a bad idea. You should find another way.

Doc, it hurts when I do this.

Then don't do that!

Upvotes: 1

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