Here first of all this will be executed. (b1=false) This is because expressions are evaluated from left to right so b1 will become false. Since this will result in a boolean value so there will be no compile time error. So the first value will be false
Then (b1&b3) will be evaluated. Since b1 is false so it will result in false.
Then (b1|b2) will be evalued. Since both b1 and b2 are false, so this will result in false.
So at last the whole condition will becode
which will result in false so beta will not be displayed.
Also your code has declarations Boolean and not boolean. Although this will work but I just want to remind you... [ August 08, 2008: Message edited by: Ankit Garg ]