Java Programming - Operators and Assignments - Discussion

Good, thanks all for explaining.

long[] a1={3,4,5};
long[] a2=a1;

This is equal to,
a1[0]=a2[0] and a1[1]=a2[1] and a1[2]=a2[2].

And in fix method, a3 is method local variable it is not an array.
So changes made to a1 is also reflected in a2 so the answer is 15 15.

It's because in java array is passed by reference, so any changes in the method reflect in the actual array.

Please tell me how to work bubble sort method with easy program? Please explain with an example.

Can anyone explain fix method in detail?

@ALL.

Refer this code.

public class PassA {

public static void main(String[] args) {
PassA p = new PassA();
p.start();
}
void start()
{
long a1[]={3,4,5};
for(long i:a1)
System.out.print(i+"" );
System.out.println(" ");
long a2[]=fix(a1);
System.out.print(a1[0]+a1[1]+a1[2]+" ");
System.out.println(a2[0]+a2[1]+a2[2]);
}
long [] fix(long [] a1)
{
a1[1] = 7;
for(long i:a1)
System.out.print(i);
System.out.println(" ");

return a1;
}
}

Output
345
375
15 15

How the value of A1[1] has changed? explain.

long[] fix(long[] a3) { // a3=a1 and a1= {3,4,5}
a3[1] = 7; // a3[1]=7 means a1[1] will become 7(a1[1]=7), now a1={3,7,5}
return a3;// return a1
}

fix() is from which package? Please explain.

According to me,

class PassA
{
public static void main(String [] args)
{
PassA p = new PassA();
p.start();
}

void start()
{
long [] a1 = {3,4,5};
long [] a2 = fix(a1);// In this line we are sending the reference variable of a1.
System.out.print(a1[0] + a1[1] + a1[2] + " ");
System.out.println(a2[0] + a2[1] + a2[2]);
}

long [] fix(long [] a3) // Now a3 also starts pointing to the same array as a1 is pointing.
{
a3[1] = 7;// Any modification done now will reflect on both a1..
return a3;//After returning this value array a2 starts will start to point a1 array as well.
}
}