The error shows this line
if ((a[0] & 1 == 0) && (a[1] & 1== 0) && (a[2] & 1== 0)){
This is the whole code:
public class Ex4 {
public static void main(String[] args) {
int [] a = new int [3];
if(args.length == 3)
{
try{
for(int i = 0; i < args.length; i++)
{
a[i] = Integer.parseInt(args[i]);
}
}
catch(NumberFormatException e){
System.out.println("Wrong Argument");
}
if ((a[0] & 1 == 0) && (a[1] & 1== 0) && (a[2] & 1== 0)){
System.out.println("yes");
}
else {
System.out.println("no");
}
}
else{
System.out.println("Error");
}
}
}
I've fixed the code:
if ((a[0] & 1) == 0 && (a[1] & 1) == 0 && (a[2] & 1) == 0){
Was an issue with the brackets, updated for anyone in the future.
== has higher precedence than &. You might want to wrap your operations in () to specify how you want your operands to bind to the operators.
((a[0] & 1) == 0)
Similarly for all parts of the if condition.
You have to be more precise, using parentheses, otherwise Java will not use the order of operands that you want it to use.
if ((a[0] & 1 == 0) && (a[1] & 1== 0) && (a[2] & 1== 0)){
Becomes
if (((a[0] & 1) == 0) && ((a[1] & 1) == 0) && ((a[2] & 1) == 0)){
Because & has a lesser priority than ==.
Your code is equivalent to a[0] & (1 == 0), and unless a[0] is a boolean this won't compile...
You need to:
(a[0] & 1) == 0
etc etc.
(yes, Java does hava a boolean & operator -- a non shortcut logical and)
Source: Stackoverflow.com