confusion on casting

[Michael Rocha]

Hi,

The following code works

class trythis {
static final short s1=10;
public static void main(String args[]) {
short s2;

s2 = s1 + 11;
System.out.println(s2);
}
}

but the following code doesn't compile

class trythis {
public static void main(String args[]) {
short s2;
        short s1=10;

s2 = s1 + 11;
System.out.println(s2);
}
}

Any ideas on why this happens?
Thanks,
Mike

[Alpana Singh]

In your first program s1 is compile time constant and therefore s2=s1+11 doesn't give an error.

While in the second one s1 is not compile time constant and hence when u add s1+11 it will result in an int which is not implicitly casted to short.

[Michael Rocha]

Thanks Alpana,

Now another related question.

 
class trythis {
public static void main(String args[]) {
short s2;
                short s1=10;

s2 = s1 + (short) 11;

System.out.println(s2);
}
}

Why doesn't this work?

Thanks,
Mike

[A Kumar]

HI..

From Aplana's post

While in the second one s1 is not compile time constant and hence when u add s1+11 it will result in an int which is not implicitly casted to short.

s2 = s1 + (short) 11;

Again here ...the result is not getting casted to short.

try this..

s2 = (short)(s1 + 11);

Regards

[ November 08, 2005: Message edited by: A Kumar ]
[ November 08, 2005: Message edited by: A Kumar ]

[Michael Rocha]

I know I'm being a bit sticky on this but I just want to clear my concepts. Please bear with me...

Doesn't the statement

    s2 = s1 + (short) 11;

imply that everything is a short?

Thanks,
Mike

[Alpana Singh]

"The result of an expression involving anything int-sized or smaller is always int".

eg.
byte b=3;
byte c=4;
byte d= b+c; //Error
byte d1=byte(b+c);//compiles fine.

Same is the case with short.

now if you code like

short s1= 10 + 11;//compiles fine

as 10 & 11 are compile time constant.

but

short s1=10;
shot s2= s1 + short(11); //here s1 is not compile time constant.it's value can change at runtime.

[Purujit Saha]

Hi,
When ever you do any aithmatic operation Result Types of Arithmetic Operations are converted to the bigger primitive.

say, x + y = r

Here operand of the arithmatic operation x & y;

Now the result will be the primitive type among x & y which ever is the largest;

In your example,

class trythis {static final short s1=10;public static void main(String args[]) {short s2;s2 = s1 + 11;System.out.println(s2);}}

The operand of arithmatic operation is short s1 & int 11;
so the result will be int type, so you need to explicitly convert int to short.
so you need to write
s2 = (short)(s1 + 11);

For details,
visit this URL

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/arithmetic.html

[Michael Rocha]

Thanks Alpana and A Kumar,

That really cleared things up for me.

Regards,
Mike

[A Enter]

hi,all
Look the follow code,It can complie correctly.

class trythis {
static final short s1=10;

public static void main(String args[]) {
short s2;
s2 = (int)s1 + 11;
System.out.println(s2);
}
}

what's the "static final" do?

[A Enter]

I think the final short(any var) can't be casting to int.
and in this case system cast int to short.
hehe.

[Mahendar Reddy]

When ever you do any aithmatic operation Result Types of Arithmetic Operations are converted to the bigger primitive.

say, x + y = r

Here operand of the arithmatic operation x & y;

Now the result will be the primitive type among x & y which ever is the largest;

Hi Purujith,I hope u know this...But remember once more...
"The result of arith operation is not ALWAYS the largest of two operands.."