Get rid of ugly if statements

Question

I have this ugly code:

if ( v > 10 ) size = 6;
if ( v > 22 ) size = 5;
if ( v > 51 ) size = 4;
if ( v > 68 ) size = 3;
if ( v > 117 ) size = 2;
if ( v > 145 ) size = 1;
return size;

How can I get rid of the multiple if statements?

Answer 1

How about such approach:

int getSize(int v) {
    int[] thresholds = {145, 117, 68, 51, 22, 10};

    for (int i = 0; i < thresholds.length; i++) {
        if (v > thresholds[i]) return i+1;
    }
    return 1;
}

---

Functionally: (Demonstrated in Scala)

def getSize(v: Int): Int = {
  val thresholds = Vector(145, 117, 68, 51, 22, 10)
  thresholds.zipWithIndex.find(v > _._1).map(_._2).getOrElse(0) + 1
}

Answer 2

Just for completeness, let me suggest that you could set up an array SIZES with 145 elements so the answer could be returned directly as SIZES[v]. Pardon me for not writing the whole thing out. You would have to make sure v was in range, of course.

The only reason I can think of for doing it that way would be if you were going to create the array once and use it thousands of time in an application that had to be really fast. I mention it as an example of a trade-off between memory and speed (not the problem it once was), and also between setup time and speed.

Answer 3

I have one more version for you. I don't really think it's the best one because it adds unnecessary complexity in the name of "performance" when I'm 100% sure this function will never be a performance hog (unless someone is calculating size in a tight loop a million times ...).

But I present it just because I thought performing a hard-coded binary search to be sort of interesting. It doesn't look very binary-y because there aren't enough elements to go very deep, but it does have the virtue that it returns a result in no more than 3 tests rather than 6 as in the original post. The return statements are also in order by size which would help with understanding and/or modification.

if (v > 68) {
   if (v > 145) {
      return 1
   } else if (v > 117) {
      return 2;
   } else {
      return 3;
   }
} else {
   if (v > 51) {
      return 4;
   } else if (v > 22) {
      return 5;
   } else {
      return 6;
   }
}

Answer 4

            if (v <= 10)
                return size;
            else {
                size = 1;

                if (v > 145)
                    return size;
                else if (v > 117)
                    return ++size;
                else if (v > 68)
                    return (size+2);
                else if (v > 51)
                    return (size+3);
                else if (v > 22)
                    return (size+4);
                else if (v > 10)
                    return (size+5);
            }

This will execute the necessary if statements only.

Answer 5

Here's my shot at it...

Update: Fixed. Previous Solution gave incorrect answers for exact values (10,22,51...). This one defaults to 6 for the if val < 10

   static int Foo(int val)
    {
                          //6, 5, 4, 3, 2 ,1
        int[] v = new int[]{10,22,51,68,117,145};
        int pos = Arrays.binarySearch(v, val-1);
        if ( pos < 0) pos = ~pos;
        if ( pos > 0) pos --;
        return 6-pos;
    }

Answer 6

Using the NavigableMap API :

NavigableMap<Integer, Integer> s = new TreeMap<Integer, Integer>();
s.put(10, 6);
s.put(22, 5);
s.put(51, 4);
s.put(68, 3);
s.put(117, 2);
s.put(145, 1);

return s.lowerEntry(v).getValue();