I am currently taking a introductory course in Java and this is regarding try-catch method. When I type this my System.out.println statement keeps repeating endlessly. Here is my code:
public static double exp(double b, int c) { if (c == 0) { return 1; } // c > 0 if (c % 2 == 0) { return exp(b*b, c / 2); } if (c<0){ try{ throw new ArithmeticException(); } catch (ArithmeticException e) { System.out.println("yadonegoofed"); } } // c is odd and > 0 return b * exp(b, c-1); } 
if (c<0){ try{ throw new ArithmeticException(); } catch (ArithmeticException e) { System.out.println("yadonegoofed"); } } // c is odd and > 0 return b * exp(b, c-1); Your comment c is odd and > 0 is incorrect -- you never actually terminated the function with the exception. You threw it, you immediately caught it, and then continued to execute the recursive function. Eventually, when you hit wraparound, it'll be a positive number again, and the errors won't happen. (It's about two billion iterations away -- don't wait.)
I would not use an exception here -- you just need to terminate the recursion. I'd check for negative input before checking for 0, and throw the exception there, and catch the exception in the caller.
In pseudo-code:
exp(double b, int c) { if (c < 0) throw new Exception("C cannot be negative"); } else if (c % 2 == 0) { return exp(b*b, c / 2); } else { /* and so forth */ } } 
You forgot one very important part when it comes to creating your own custom exceptions. You forgot to tell the method that it will throw such a method. Your first line of code should look like this:
public static double exp(double b, int c) throws ArithmeticException { Note that I have tested this myself, and it will only throw the exception one time in your output.
If for example, c = -1 in, the first and second if fails, the third if throws an exception and then prints the error, but things progress because you handled the excpetion. So it calls exp(b, -2). In turn, that calls exp(b, -3) in the return, and so on. Add the value of C to your println to verify.
println to verify.. That will point out the problem very quickly. :)Well, right at the end you have return b * exp(b, c-1); this is going to call exp again which will call it again.
So the function will keep repeating and so will System.out.println.
Your BASE case is VERY specific...what in your code GUARANTEES that c will equal 0? Currently, this is the ONLY way that you exit your recursive call. As @Jay said you always subtract 1 which causes the thrown exception, but c is already below 0 at that point so it doesn't EQUAL 0. Change your first if statement to catch values <= 0 and you should be fine.
if( c <= 0 ) return 1; c < 0 is probably an error, and should be handled as an error.