Too many 'if' statements?

Question

The following code does work how I need it to, but it\'s ugly, excessive or a number of other things. I\'ve looked at formulas and attempted to write a few solutions, but I

Answer 1

If you cannot come up with a formula, you can use a table for such a limited number of outcomes:

final int[][] result = new int[][] {
  { 0, 0, 1, 2 },
  { 0, 0, 2, 1 },
  { 2, 1, 3, 3 },
  { 1, 2, 3, 3 }
};
return result[one][two];

Answer 2

Other people have already suggested my initial idea, the matrix method, but in addition to consolidating the if statements you can avoid some of what you have by making sure the arguments supplied are in the expected range and by using in-place returns (some coding standards I've seen enforce one-point-of-exit for functions, but I've found that multiple returns are very useful for avoiding arrow coding and with the prevalence of exceptions in Java there's not much point in strictly enforcing such a rule anyway as any uncaught exception thrown inside the method is a possible point of exit anyway). Nesting switch statements is a possibility, but for the small range of values you're checking here I find if statements to be more compact and not likely to result in much of a performance difference, especially if your program is turn-based rather than real-time.

public int fightMath(int one, int two) {
    if (one > 3 || one < 0 || two > 3 || two < 0) {
        throw new IllegalArgumentException("Result is undefined for arguments outside the range [0, 3]");
    }

    if (one <= 1) {
        if (two <= 1) return 0;
        if (two - one == 2) return 1;
        return 2; // two can only be 3 here, no need for an explicit conditional
    }

    // one >= 2
    if (two >= 2) return 3;
    if (two == 1) return 1;
    return 2; // two can only be 0 here
}

This does end up being less readable than it might otherwise be due to the irregularity of parts of the input->result mapping. I favor the matrix style instead due to its simplicity and how you can set up the matrix to make sense visually (though that is in part influenced by my memories of Karnaugh maps):

int[][] results = {{0, 0, 1, 2},
                   {0, 0, 2, 1},
                   {2, 1, 3, 3},
                   {2, 1, 3, 3}};

---

Update: Given your mention of blocking/hitting, here's a more radical change to the function that utilizes propertied/attribute-holding enumerated types for inputs and the result and also modifies the result a little to account for blocking, which should result in a more readable function.

enum MoveType {
    ATTACK,
    BLOCK;
}

enum MoveHeight {
    HIGH,
    LOW;
}

enum Move {
    // Enum members can have properties/attributes/data members of their own
    ATTACK_HIGH(MoveType.ATTACK, MoveHeight.HIGH),
    ATTACK_LOW(MoveType.ATTACK, MoveHeight.LOW),
    BLOCK_HIGH(MoveType.BLOCK, MoveHeight.HIGH),
    BLOCK_LOW(MoveType.BLOCK, MoveHeight.LOW);

    public final MoveType type;
    public final MoveHeight height;

    private Move(MoveType type, MoveHeight height) {
        this.type = type;
        this.height = height;
    }

    /** Makes the attack checks later on simpler. */
    public boolean isAttack() {
        return this.type == MoveType.ATTACK;
    }
}

enum LandedHit {
    NEITHER,
    PLAYER_ONE,
    PLAYER_TWO,
    BOTH;
}

LandedHit fightMath(Move one, Move two) {
    // One is an attack, the other is a block
    if (one.type != two.type) {
        // attack at some height gets blocked by block at same height
        if (one.height == two.height) return LandedHit.NEITHER;

        // Either player 1 attacked or player 2 attacked; whoever did
        // lands a hit
        if (one.isAttack()) return LandedHit.PLAYER_ONE;
        return LandedHit.PLAYER_TWO;
    }

    // both attack
    if (one.isAttack()) return LandedHit.BOTH;

    // both block
    return LandedHit.NEITHER;
}

You don't even have to change the function itself if you want to add blocks/attacks of more heights, just the enums; adding additional types of moves will probably require modification of the function, though. Also, EnumSets might be more extensible than using extra enums as properties of the main enum, e.g. EnumSet<Move> attacks = EnumSet.of(Move.ATTACK_HIGH, Move.ATTACK_LOW, ...); and then attacks.contains(move) rather than move.type == MoveType.ATTACK, though using EnumSets will probably be slightly slower than direct equals checks.

---

For the case where a successful block results in a counter, you can replace if (one.height == two.height) return LandedHit.NEITHER; with

if (one.height == two.height) {
    // Successful block results in a counter against the attacker
    if (one.isAttack()) return LandedHit.PLAYER_TWO;
    return LandedHit.PLAYER_ONE;
}

Also, replacing some of the if statements with usage of the ternary operator (boolean_expression ? result_if_true : result_if_false) could make the code more compact (for example, the code in the preceding block would become return one.isAttack() ? LandedHit.PLAYER_TWO : LandedHit.PLAYER_ONE;), but that can lead to harder-to-read oneliners so I wouldn't recommend it for more complex branching.

Answer 3

The first thing that occurred to me was essentially the same answer given by Francisco Presencia, but optimized somewhat:

public int fightMath(int one, int two)
{
    switch (one*10 + two)
    {
    case  0:
    case  1:
    case 10:
    case 11:
        return 0;
    case  2:
    case 13:
    case 21:
    case 30:
        return 1;
    case  3:
    case 12:
    case 20:
    case 31:
        return 2;
    case 22:
    case 23:
    case 32:
    case 33:
        return 3;
    }
}

You could further optimize it by making the last case (for 3) the default case:

    //case 22:
    //case 23:
    //case 32:
    //case 33:
    default:
        return 3;

The advantage of this method is that it is easier to see which values for one and two correspond to which return values than some of the other suggested methods.

Answer 4

I don't like any of the solutions presented except for JAB's. None of the others make it easy to read the code and understand what is being computed.

Here's how I would write this code -- I only know C#, not Java, but you get the picture:

const bool t = true;
const bool f = false;
static readonly bool[,] attackResult = {
    { f, f, t, f }, 
    { f, f, f, t },
    { f, t, t, t },
    { t, f, t, t }
};
[Flags] enum HitResult 
{ 
    Neither = 0,
    PlayerOne = 1,
    PlayerTwo = 2,
    Both = PlayerOne | PlayerTwo
}
static HitResult ResolveAttack(int one, int two)
{
    return 
        (attackResult[one, two] ? HitResult.PlayerOne : HitResult.Neither) | 
        (attackResult[two, one] ? HitResult.PlayerTwo : HitResult.Neither);
}    

Now it is much more clear what is being computed here: this emphasizes that we are computing who gets hit by what attack, and returning both results.

However this could be even better; that Boolean array is somewhat opaque. I like the table lookup approach but I would be inclined to write it in such a way that made it clear what the intended game semantics were. That is, rather than "an attack of zero and a defense of one results in no hit", instead find a way to make the code more clearly imply "a low kick attack and a low block defense results in no hit". Make the code reflect the business logic of the game.

Answer 5

Let's see what we know

1: your answers are symmetrical for P1 (player one) and P2 (player two). This makes sense for a fighting game but is also something you can take advantage of to improve your logic.

2: 3 beats 0 beats 2 beats 1 beats 3. The only cases not covered by these cases are combinations of 0 vs 1 and 2 vs 3. To put it another way the unique victory table looks like this: 0 beats 2, 1 beats 3, 2 beats 1, 3 beats 0.

3: If 0/1 go up against each other then there's a hitless draw but if 2/3 go up against each then both hit

First, let us build a one-way function telling us if we won:

// returns whether we beat our opponent
public boolean doesBeat(int attacker, int defender) {
  int[] beats = {2, 3, 1, 0};
  return defender == beats[attacker];
}

We can then use this function to compose the final result:

// returns the overall fight result
// bit 0 = one hits
// bit 1 = two hits
public int fightMath(int one, int two)
{
  // Check to see whether either has an outright winning combo
  if (doesBeat(one, two))
    return 1;

  if (doesBeat(two, one))
    return 2;

  // If both have 0/1 then its hitless draw but if both have 2/3 then they both hit.
  // We can check this by seeing whether the second bit is set and we need only check
  // one's value as combinations where they don't both have 0/1 or 2/3 have already
  // been dealt with 
  return (one & 2) ? 3 : 0;
}

While this is arguably more complex and probably slower than the table lookup offered in many answers I believe it is a superior method because it actually encapsulates the logic of your code and describes it to anyone who's reading your code. I think this makes it a better implementation.

(It's been a while since I did any Java so apologies if the syntax is off, hopefully it is still intelligible if I've got it slightly wrong)

By the way, 0-3 clearly mean something; they're not arbitrary values so it would help to name them.

Answer 6

((two&2)*(1+((one^two)&1))+(one&2)*(2-((one^two)&1)))/2