Assign bitwise operators in to variable in java

Question

I try to assign bit wise operators into variables. How can i do it?

Eg

    String bitOp="";

        String rFunction=JOptionPane.showInputDialog(null,"Enter Your round function","Round function",JOptionPane.INFORMATION_MESSAGE);

         if(rFunction.toUpperCase()=="AND")
             bitOp="&";
         if(rFunction.toUpperCase()=="OR")
             bitOp="|";
         if(rFunction.toUpperCase()=="XOR")
             bitOp="^";

    int res= x +bitOp+ y; // if the operator is "AND" , this should be x & y.
                          //Here NumberFormatException error shows.
    System.out.print(res);

But it doesn't work. Anyone please!!!

You are essentially trying to do "eval." Take a look at [this](http://stackoverflow.com/questions/2605032/is-there-an-eval-function-in-java). — intboolstring, Jun 12 '16 at 05:35
Also, you'll need to perform the calculation (not assign a "bitOp"). — Elliott Frisch, Jun 12 '16 at 05:36
@ElliottFrisch that is the wrong duplicate. Even if he had the string comparison correct, that wouldn't be the problem. — intboolstring, Jun 12 '16 at 05:36
You don't need to write `rFunction.toUpperCase()=="AND"` just simply do `rFunction.equalsIgnoreCase("AND");` — Shree Krishna, Jun 12 '16 at 05:37
@RolandIllig could u please tell me how to use it in above code? — , Jun 12 '16 at 05:50
@RolandIllig That's right but He is using `.toUpperCase()` and comparing the String, so that I suggested to use `equalsIgnoreCase` instead — Shree Krishna, Jun 12 '16 at 05:52
`int res= x +bitOp+ y;` the problem is here, it shows NumberFormatException. Because i defined them as string — , Jun 12 '16 at 05:56
*it shows NumberFormatException* This most certainly does not show a `NumberFormatException`. — Tom, Jun 12 '16 at 06:16

Turo · Answer 1 · 2016-06-12T08:36:55.397

You could make your own BitOperator:

public enum BitOperator {
    AND {
        @Override
        public int calc(int x, int y) {
            return x & y;
        }
    },
    OR {
        @Override
        public int calc(int x, int y) {
            return x | y;
        }
    },
    XOR {
        @Override
        public int calc(int x, int y) {
            return x ^ y;
        }
    };

    public abstract int calc(int x,int y);

}

Usage(add some Exception handling):

          String rFunction=JOptionPane.showInputDialog(null,"Enter Your round function","Round function",JOptionPane.INFORMATION_MESSAGE);
          System.out.println(BitOperator.valueOf(rFunction.toUpperCase()).calc(x, y));

score 0 · Answer 2 · answered Jun 12 '16 at 08:21

Your code shouldn't even compile so I am not sure of what you really want to achieve. Do you want to evaluate the String expression you create or do you just want to get the result more classically? Where do x and y come from...? There is an important part of your code that is missing.

Anyway I would chose the second option:

int eval;

String rFunction=JOptionPane.showInputDialog(null,"Enter Your round function","Round function",JOptionPane.INFORMATION_MESSAGE).toUpperCase();

 if(rFunction.equals("AND")){
     eval = x & y;
 } else if(rFunction.equals("OR")){
     eval = x | y;
 } else if(rFunction.equals("XOR")){
     eval = x ^ y;
 } else {
     //here you could also throw an exception
     //or loop and request the user to renew their choice
     System.out.print("Invalid choice");
     return;
 }

System.out.print(res);

Assign bitwise operators in to variable in java

2 Answers2