Doubt on byte literals

[bhanu chowdary]

Doing some code snippets during my preparation and i encountered this. The below code throws a compiler error

 public class LiteralsTest {
public static void main(String[] args) {
byte s = 10;
byte s1 = 10;
byte s2 = s+s1; //cannot convert from int to byte
}
} 

but if i put the modifier final to both s and s1 it is compiling.

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

Am i missing something here? why is the compiler not throwing an error?

[Devaka Cooray]

Please check this one.

[Prithvi Sehgal]

Hello,

It is because i believe this conversion is taking place at compile time rather then at run-time as variables are declared final.
As they are within the range of the byte, that is why it is not giving any compilation error. Lets say

final byte s = 10;
final byte s1 = 120;
byte s2 = s+s1;
System.out.println(s2);

It will start giving compile time error as it has gone out of range. But above conversion i suppose is happening at compile
time rather then at run-time. I would love if someone else will comment.

Am i right or not.

Best Regards,

[Ankit Garg]

In the second case, both s and s1 are compile time constants. This means that the compiler knows their value at compile time. So when you say s+s1, the compiler is able to find the result of that expression i.e. the compiler knows that s+s1 will be 20, so it replaces the expression by its value. So your code after compilation becomes

public class LiteralsTest {
public static void main(String[] args) {
final byte s = 10;
final byte s1 = 10;
byte s2 = 20; //value replaces the expression
System.out.println(s2);
}
}

[bhanu chowdary]

I got it now. Thank you Devaka, Prithvi, Ankit for your replies.

[Prithvi Sehgal]

You are welcome Bhanu.

Cheers,