Incremeant - Marcus Greens Exam-3, Q-54

[rehan hamid]

public class Inc{
public static void main(String argv[]){
Inc inc = new Inc();
int i =0;
inc.fermin(i);
System.out.println(
i = i++;
System.out.println(i);
}
void fermin(int i){
i++;
}
}
Can any body tell me how output is 0. I think after execution of (i = i++) i must be 1.
Rehan.

[lee dalais]

here's a discussion
http://forum.java.sun.com/read/16789542/qAa6ifNJoxK4AAotL#LR

[sandeep bagati]

i=i++; will be 0 since first the post increment operator(++) on i is evaluated which after the statement would have changed the value to 1 and then the assignment of old value of i to i is done keeping it 0;
hence i maintains the value 0;

[rehan hamid]

Thanks I got it!!

[ryan burgdorfer]

but in this case, the method "fermin" is called on i first...incrementing it. So shouldn't the output be 1?
1. first i = 0
2. inc.fermin(i); causes i to increment to 1
3. i=i++ still keeps i at one when printed
If the method were not called, then i would be 0, but it is...
I am confused. I will compile and run the code to see what happens...

[ryan burgdorfer]

// increment test
public class Inc
{
    public static void main( String[] args )
    {
        Inc inc = new Inc() ;
        int i = 0 ;
        inc.fermin( i ) ;
        
        System.out.println( i = i++ ) ;
        System.out.println( i ) ;
    }
    
    void fermin( int i )
    {
        i++ ;
    }
}

Output is indeed
0
0
This confuses me, since the method fermin(i) is called before the print statements...
[This message has been edited by ryan burgdorfer (edited March 02, 2001).]
[This message has been edited by ryan burgdorfer (edited March 02, 2001).]

[Rajesh R]

The method fermin increment the formal paramter i not the local variable declared i in main. The scope of the formal parameter is the entire body of the method. Outside the scope it does not exist.

[Sivalingam Sivasuthan]

Hi ryan burgdorfer:
1) we are declaring and initializing a Local variable int i to zero.
2) Then we are passing that i to a method, which is incrementing, that i. Here we should understand that copy of value is passed so the original variable remain unchanged.
3) Then we print the result of the expression I=I++; that is zero.

1) The expression is having two operations i = i , i = i + 1
2) Substitute value of i in each i = 0 , i = 0 + 1
3) evaluate each i = 0 , i = 1
4) for assignment operation, associavity is Right to left so set i = 1 and then set i = 0
5) So the final result will be i = 0 

4)At the last we are printin the value of the I which is zero
Hope this helps.
Siva

[Anonymous]

Well put Siva! It would be different if the fermin method had a return, like so:
import java.io.*;
import UU.*;
// increment test
public class Inc
{
public static void main( String[] args )
{
Inc inc = new Inc() ;
int i = 0 ;
i = inc.fermin( i ) ;

System.out.println( i = i++ ) ;
System.out.println( i ) ;
}

int fermin( int i )
{
i++ ;
return i ;
}
}
Output:
1
1
Process Exit...

[ryan burgdorfer]

<*LIGHT BULB*>
Thanks guys for the enlightenment.

[Marcus Green]

This is the answer I give to that question.. (by the way can anyone confirm if you are likely to get a question on the way var++ works?).

"The method fermin only receives a copy of the variable i and any modifications to it are not reflected in the version in the calling method. The post increment operator ++ effectivly modifes the value of i after the initial value has
been assiged to the left hand side of the equals operator. This can be a very tricky conept to understand"
(hmm just noticed that spelling error conept better fix it).
Marcus