递归方法堆栈溢出错误

Question

In my code i am just trying to make a simple program that tells you if one number can divide into another number evenly (in this case that number is 3). right now I am saying that if x (the number doesnt divide evenly add 0.01 to it, that gives me the stack overflow error. If I make the value 0.2 it says that 9 is a divisible of three when really the next thing number that divides into three after three is 6

public class divisible {

   public static void divide(double x) {
      double three = 3;
      double value = x%three;

      if (value==0) {
         System.out.println(x + " is a divisible of 3 ");
         return;
      }else{ 
         //System.out.println("x does not divide evenly into 3");
         divide(x+(.01));
      }

   }

   public static void main(String args[]) {
      divide(4);
   }
}

Answer 1

The reason the recursion is infinite is a little obscure: the 0.1 cannot be represented exactly in double . When you add 0.1 to 3 ten times, you do not get a 4 - you get a number that's close to 4 , but a little greater than it. That number does not divide your target evenly, so the recursion goes on to 4.1 , 4.2 , and so on, with no end.

The reason is the same why this loop never stops ( try it! ):

for (double x = 3 ; x != 4 ; x += 0.1) {
    System.out.println(x);
}

Using BigDecimal in place of double would fix the problem, because 0.1 would be represented exactly. Of course it would still print a wrong message (the "is divisible of 3" is hard-coded, even though x could be an entirely different number in the invocation when the remainder becomes zero).

Answer 2

Your problem is that you compare two doubles with == . This will produce unreliable results, due to the way that floating point arithmetic is implemented. Your method should look like this:

public static void divide(int x) {
  int three = 3;
  int value = x%three;

  if (value==0) {
     System.out.println(x + " is a divisible of 3 ");
     return;
  }else{ 
     System.out.println("x does not divide evenly into 3");
//         divide(x+(.01));
  }

}

If you want the method to be accessed with a double argument, you can cast:

public static void divide(double x) {
    x = (int) Math.round(x);

If you want to be able to handle number larger than Integer.MAX_VALUE , you can use BigInteger