initialization and if

[Jini Varghese]

What would happen when the following is compiled and executed. Select the one correct answer
public class compare {
public static void main(String args[]) {
int X=10,y;
if(X<10) y=1;<br /> if(X>=10) y=2;
System.out.println("y is "+y);
}
}
a. the program compiles and prints y is 0 when executed
b. the program compiles and prints y is 1 when executed
c. the program compiles and prints y is 2 when executed
d. the program does not compile complaining about y not being initialized
e. the program throws a runtime exception
ans :d
I did the code and the answer is d. But y is given value for all (+ve and -ve) values of X. So why compiler complains? Is there anything that I am missing here?
Thanks in advance...

[Ravindra Mohan]

Hi Jini,

public class compare {
public static void main(String args[]) {
int X=10,y;
if(X<10) y=1; // the problem lies here
              // as the compiler identified
              // that y will not be initialised in case "x" is >= 10. 
             // hence it complains     
if(X>=10) y=2;
System.out.println("y is "+y);
} 
}

Instead if you coded something like

public class compare {
public static void main(String args[]) {
int X=10,y;
if(X<10) y=1; 
else
y= 2;
System.out.println("Y =" +y);
}
}  

Then the compiler does realise that variable "y" will definately be initialised no matter whatever the condition is evaluated in the if block .
Hope this clears your doubts.
Ravindra Mohan.

[This message has been edited by Ravindra Mohan (edited May 10, 2001).]

[Art Metzer]

Hi, Jini.
Java is smart to enough to figure out mutual exclusivity in this case:

<font size="2.0">
public class Compare {
    public static void main( String args[] ) {
        int x = 10, y;
        if ( x < 10 ) {
            y = 1;
        }
        else {
            y = 2;
        }
        System.out.println( "y is " + y );
    }
}

but not in your case:

</font size="2.0">
public class Compare {
    public static void main( String args[] ) {
        int x = 10, y;
        if ( x < 10 ) {
            y = 1;
        }
        if ( x >= 10 ) {
            y = 2;
        }
        System.out.println( "y is " + y );
    }
}

Art

[Micah Holroyd]

If you make the X variable final (see code below), then the compiler knows that X will always be 10 and can figure out that Y will be initialized. In the original code, the compiler can't be absolutely sure of what the value of X will be, so it can't predict whether Y will be initialized or not.

public class compare {
public static void main(String args[]) {
final int X=10;
int y;
if(X<10) y=1;
if(X>=10) y=2;
System.out.println("y is "+y);
}
}

[Jini Varghese]

Great...That really helped. Thanks a lot.

[Samith Nambiar]

hi micah

i tried to compile your code with the final modifier but it still does give me a compilation error
regards
Samith.P.Nambiar

[Jini Varghese]

The second one, with final modifier still gives error.