the operator type & is undefined for the argument type(s) int - What gves?

Grumpy_Mike

The reference page says that the bitwise AND operator & can be used with variable type int, but when you try it you get an error. Any idea what is going on? Is it Processing, the reference page or me? Processing 3.3.3 on a Mac Thanks

koogs

Post the statement giving the error.

Grumpy_Mike

int num = 0xE6;
 if(num & 0x01 == 0)
 { print("fine");
 }

This works in C

Chrisir

maybe like this? I dunno.

In your version your & might be treated as &&

    int num = 0xE6;
    if ((num & 0x01) == 0)
    { 
      print("fine");
    }

quark

The problem appears to be that Processing has evaluated the 0x01 == 0 first and then tries to do the bitwise & between an int and a boolean. The solution is to use parentheses to clarify the operedr of evaluation like this

int num = 0xE6;
if ((num & 0x01) == 0)
{ 
  print("fine");
}

GoToLoop

Like every1 said above. Just an extra info:
& 1 is used to determine whether a value is even or odd: B-)

int val = (int) random(10);
boolean isEven = (val & 1) == 0;
println(val, isEven);
exit();

GoToLoop

Another curious case in Java: Operators &, | and ^ are overloaded to accept boolean operands:

if (mouseX > width>>2 & mouseX < 3*width/4) println("Active region"); :)>-

However, both sides of the bitwise operators are always eagerly evaluated. L-)

Unlike their logical counterparts && and ||, which are lazy-evaluated and short-circuits the right side if the result can already be determined by just the left side operand. :-B

Grumpy_Mike

Thanks guys that did it.

koogs

the actual error message is

"The operator & is undefined for the argument type(s) int, boolean"

things are so much easier when people give the exact error message.