Pass by value

[Nakul P. Patel]

Example of pass by value:
public class Passbyvalue {

public static void main(String[] args) {

int a=15,b;

System.out.println("Value of a before call:"+a);
Value value = new Value();
value.meth(a);
System.out.println("Value of a after call:"+a);

}
}


public class Value {

void meth(int i){
i *= i;

}
}
Here the value of a remains as 15.
If i perform same operation with return value the value of a is changed.


public class Passbyvalue {

public static void main(String[] args) {

int a=15,b;

System.out.println("Value of a before call:"+a);
Value value = new Value();
value.meth(a);

b= value.meth(a);
System.out.println("Value of a after call:"+b);
}
}


public class Value {

int meth(int i){
i *= i;
return i;
}
}
here value of a is changed to 255.It is pass by value and it changes the value.
Can you explain it HOW?

[fred rosenberger]

Nakuli Patidar wrote:

Here the value of a remains as 15.
If i perform same operation with return value the value of a is changed.

public class Passbyvalue {
	
	public static void main(String[] args) {
	
		int a=15,b;
			
		System.out.println("Value of a before call:"+a);
		Value value = new Value();
		value.meth(a);
			
		b= value.meth(a);
		System.out.println("Value of a after call:"+b);
	}	
}


public class Value {

	int meth(int i){
		i *= i;
		return i;	
	}
}

here value of a is changed to 255.It is pass by value and it changes the value.
Can you explain it HOW?

Where do you see that a has changed? the value of b gets set to 255, but if you actually printed a (you are printing b), you'll see it doesn't change.

[Nakul P. Patel]

Even after printing value of a:

a= value.meth(a);
System.out.println("Value of a after call:"+a);

the value of a is changed.

I request you all to run this code.

Please let me know if you have any idea.

[Paul Clapham]

Of course the value of a is changed. Your code assigns it the value returned by the method, so it is completely unsurprising that it gets that value. This has nothing to do with parameter passing at all.

[James Boswell]

Nakuli

value.meth(a);

Try this instead.

[fred rosenberger]

Nakuli Patidar wrote:

a= value.meth(a);
System.out.println("Value of a after call:"+a);

you've changed what you're doing here. "a" changes because you do some computation, and then reset "a" to that new value. if you had done this:

a = 7;
System.out.println("Value of a after call:" +a);

would you be surprised that "a" changed? No. What you should have done to see that "a" didn't change was this:

b= value.meth(a); 
System.out.println("Value of a after call:"+a);