Thread: how to send a message to receiver with if else statement

I'm not really sure what you are trying to do to be honest. Do you want to pass the number of chickens on to some other method?

// Json

No I dont need it to pass it on I just need it to send a message to the receiver with the value of the argument in the message. so if the value of the argument is 1 I need it to send a message to the receiver saying "chickens1()" and if the value of the argument is 2 it should be "chickens2" and so on.

hope this makes it more clear for you.

Do you mean somthing like this?

 public void countChickens(int chickens)
      {
          if  (chickens >6)
          {
                 this.chickens("chickens" + Integer.toString(chickens));
           }
       }

Hi that looks like something that would work, however when i tried this the compiler came up with the error "cannot find symbol - method chickens(java.lang.string"

what I think this error is saying is that the compilere thinks chickens is a method??

hope someone can help me further. Thankyou for the help so far. I will be sure to thank everyone properly when the problem is resolved.

Yes, the compiler (as well as myself) thought chickens was a method, so in that case there should be a method like

public void chickens( String s ){
    //do something
}

Of course whether or not there should be a chickens() method - and if so what it does - is dependent upon the behavior you want, and that I'm a bit unclear of so if you could describe what you want your code to do a bit further it may help

Yeah this is sort of what I meant, you could just put the number of chickens into the this.chickens method which would then print whatever message is needed.

// Json

Assuming that both face1() or whatever you decide to call it, and face2() etc all have the same medthod headers then this problem can be easily solved using Function objects, since Java doesn't support Function pointers.

consider the following code:

import java.util.Scanner;
 
 
public class FP {
	public static void main(String[] args) {
		FuncO[] y = new FuncO[2];
 
		y[0] = new FuncO(){
 
			@Override
			public int override(int x) {
 
				return x*x;
			}
 
		};
 
		y[1] = new FuncO(){
 
			@Override
			public int override(int x) {
 
				return x+x;
			}
 
		};
 
		System.out.print("Enter 0 to square a number, 1 to add the number to its self:> ");
		Scanner s = new Scanner(System.in);
		System.out.println(y[s.nextInt()].override(s.nextInt()));
	}
}
 
interface FuncO{
	public int override(int x);
}

Chris

Hi Chris,

Thanks for taking the time to respond. Unfortunately, function objects have not been covered yet in the course, so I wouldn't be able to use them even if I knew how.

So far we've covered: Object concepts, variables,representations, basic methods, selection and iteration.

In that case, you still have the funxtions hard coded so....

switch(x){
   case 1: this.chickens1(); break;
   case 2: this.chickens2(); break;
   case 3: this.chickens3(); break;
   case 4: this.chickens4(); break;
   case 5: this.chickens5(); break;
   case 6: this.chickens6(); break;
   default: break;
}

Chris