1. public class TestA
2. {
3. public static void main(String[] args)
4. {
5.
6. boolean flag = true;
7. byte b = (flag==true) ? 0 : 1;
8. System.out.println(b);
9. }
10.}

Understandably , I get a compile error on line 6 :
Incompatible type for declaration. Explicit cast needed to convert int to byte.

If I change line 6 to:
byte b = (true==true) ? 0 : 1;

This compiles okay. Is it because of compiler optimisation on the modified line no 6 making it equal to:
byte b = 0;

Regds.

byte b = (flag==true) ? 0 : 1;
when we use a var(not constant) flag in an expression it is NOT COMPILE TIME CONSTANT. The variable's value may change at run time. We can't be sure. This means the right hand side of the above statement is NOT COMPILE TIME constant expression and the evaluation of the right hand side of the statement is common promoted type of the 2 operands between the : . Here both are 'int' So the compiler says Can't convert 'int' to 'byte' at compile time. IF the operands at both sides of the : WOULD HAVE BEEN int : float then the compiler will say can't convert float to byte. This rule is applicable when both operands are reference types also. One operand MUST be convertible to the other.

So, when you apply flag==true the resulting type of the above statement becomes int which CAN'T be assigned to a var of type byte.
At the same time when the WHOLE of right hand side expression involves ALL Compile time constants like true == true and the resulting int literal value is between the range of the left hand var type which here is 'byte' ,then this expression is ELIGIBLE for the above said special type of assignment conversion So the foll.
byte b = (true==true) ? 0 : 1;
same as
byte b = true ? 0 : 1;
same as
byte b= 0;

You can learn more about the terinary operator here in JLS

regds
maha anna

This is a good question. It kind of stumped me at first, and it brought out something about conditional expressions that I wasn't aware of.

What I think is happening here is that if you use a variable in the boolean part of the conditional expression, then the second and third operands are subject to (binary?) numeric promotion and an explicit cast is required.

If you use a boolean literal, then the second and third operands are treated just as they are (i.e., constants of type int whose value can fit into the byte variable and therefore no cast is required) and the conditional expression is treated as a simple assignment.

Can someone confirm?