doubt pass by copy

[xie li]

public class Inc{
public static void main(String argv[]){
Inc inc = new Inc();
int i =0;
inc.fermin(i);
i = i++;
System.out.println(i);
}
void fermin(int i){
i++;
}
}
who can explain why the assignment to var i has no effect?
thank you

[Christophe Verré]

In i = i++,
i++ returns the evaluation of i before the incrementation.
So it assigns i to i.

[xie li]

o ,i know
thank u very much

[Gregory Jones]

Hello,

I have been reading the boards for a while and now ready to start giving back. I hope to help and not add to the confusion.

I think the difference is that primitives are passed by value and objects are passed by reference. Here is some code demonstrating the object by reference.

    
class Foo {
    int bar = 0;
}
 
public class Inc{
    public Foo foo = null;
    
    public Inc() {
        foo = new Foo();
        System.out.println(foo.bar);
        fermin(foo);
        System.out.println(foo.bar);
    }
    
    public static void main(String argv[]){
        Inc inc = new Inc();
    }
 
    void fermin(Foo i){
        i.bar++;
    }
}

HTH
Greg

[vandu matcha]

yes ..here the point is that i value is passed by value to the method..so if we increment i..it is local to the method...and when it is returned i is not incremented...and also..after returning from the method..i is again post incremented so i is not incremented again..hence the value of i remained 0...

[Keith Lynn]

Originally posted by Gregory Jones:

I think the difference is that primitives are passed by value and objects are passed by reference.

Actually all parameter passing in Java is by value. When the formal parameter and actual paramters are object references, then a copy of the reference is passed to the method.