Lots of If conditions in Java

Question

I've just started learning java and have a really basic question. I have a label that I want to change colors when a random integer between 1 and 18 lands on a particular number. These numbers are not odd or even, so I can't use that.

Right now I have this:

    if (Random == 1 || Random == 2 || Random == 5 || Random == 7 || Random == 12 || Random == 14 || Random == 16 || Random == 18)  
        label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_BLUE));
    else if (Random == 3 || Random == 4 || Random == 6 || Random == 8 || || Random == 9 | Random == 10 || Random == 11 || Random == 13 || Random == 15 || Random == 17)
        label_wheelNumber.setForeground(SWTResourceManager.getColor(SWT.COLOR_GREEN));

I know it looks silly, and I feel like an idiot doing it this way. What do you recommend? I haven't taken a class so any explanations are extremely useful. Thanks

I have looked at that, but I'm not doing so well on implementing it. Do you have an example similar to my case? — user3794455, Jul 01 '14 at 15:22
I added it to my answer below, you'll need to flesh it out a bit, but the core functionality is there. — Patrick J Abare II, Jul 01 '14 at 15:28
See http://stackoverflow.com/questions/519422/what-is-the-best-way-to-replace-or-substitute-if-else-if-else-trees-in-program — Raedwald, Jul 02 '14 at 06:56

Patrick J Abare II · Answer 1 · 2014-07-01T15:30:46.813

10

Here is an example of a switch: Note the break; when using a switch, the case will fall through. Essentially, case 1: will fall through to the next code block. For example in my code, in case 5: if the break; was not there, it would fall through to the next code block and end up with the second code block containing SWT.COLOR_GREEN being called as well.

switch(Random)
{
    case 1:
    case 2:
    case 5:
        label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_BLUE));
        break;
    case 9:
    case 10:
        label_wheelNumber.setForeground(SWTResourceManager.getColor(SWT.COLOR_GREEN));
        break;
}

edited Jul 01 '14 at 15:30

answered Jul 01 '14 at 15:23

Patrick J Abare II

1,129
1
10
31

Thank you for explaining the break when using a switch! – user3794455 Jul 01 '14 at 15:32
@LuiggiMendoza Oh Sorry. Your code too. Don't know who posted before. both are showing same time. – Suresh Atta Jul 01 '14 at 15:38
@sᴜʀᴇsʜᴀᴛᴛᴀ you can hover the mouse over the time in the posts to check the specific time the answer was posted. And the difference in this case is few seconds. – Luiggi Mendoza Jul 01 '14 at 15:39
@LuiggiMendoza Not even few. Only one second difference. :) and your's is nice answer. – Suresh Atta Jul 01 '14 at 15:41

Luiggi Mendoza · Accepted Answer · 2014-07-01T15:42:05.630

You may use a switch:

switch(Random) {
    case 1: case 2: case 5: case 7: case 12: case 14: case 16: case 18:
        //something...
        break;
    case 3: case 4: case 6: case 8: case 9: case 10: case 11: case 13: case 15: case 17:
        //something...
        break;
    default: 
        //just in case none of the over values was selected
}

If the values may vary fast or you want to allow more values, you can store them in an array or similar:

static final int[] FOREGROUND_BLUE = {1, 2, 5, 7, 12, 14, 16, 18};
static final int[] FOREGROUND_GREEN = {3, 4, 6, 8, 9, 10, 11, 13, 15, 17};

And then perform a search to seek if the value belongs to the specified array:

//using binary search since the data in the array is already sorted
int found = Arrays.binarySearch(FOREGROUND_BLUE, Random);
if (found >= 0) {
    //something...
}
found = Arrays.binarySearch(FOREGROUND_GREEN, Random);
if (found >= 0) {
    //something...
} else {
    //...
}

In case you can even have more options, probably you want to use a cache-like approach and store the data in a Map<Integer, Color>:

static final Map<Integer, Color> colorMap;
static {
    Map<Integer, Color> colorMapData = new HashMap<Integer, Color>();
    Color blue = SWTResourceManager.getColor(SWT.COLOR_BLUE);
    Color green = SWTResourceManager.getColor(SWT.COLOR_GREEN);
    colorMapData.put(1, blue);
    colorMapData.put(2, blue);
    colorMapData.put(3, green);
    colorMapData.put(4, green);
    colorMapData.put(5, blue);
    //...
    //this makes colorMap truly constant and its values cannot be modified
    colorMap = Collections.unmodifiableMap(colorMapData);
}

And then you just call the value from map:

Color = colorMap.get(Random);

Can you not use boolean operators? like `case 1 || 2:` etc.? — jhobbie, Jul 01 '14 at 15:24
@jhobbie not in a `switch` but this example emulates the `||` in your `if`s. — Luiggi Mendoza, Jul 01 '14 at 15:24

score 1 · Answer 3 · answered Jul 01 '14 at 15:22

Use a Switch statement:

public class SwitchDemo {
   public static void main(String[] args) {

    int month = 8;
    String monthString;
    switch (month) {
        case 1:  monthString = "January";
                 break;
        case 2:  monthString = "February";
                 break;
        case 3:  monthString = "March";
                 break;
        case 4:  monthString = "April";
                 break;
        case 5:  monthString = "May";
                 break;
        case 6:  monthString = "June";
                 break;
        case 7:  monthString = "July";
                 break;
        case 8:  monthString = "August";
                 break;
        case 9:  monthString = "September";
                 break;
        case 10: monthString = "October";
                 break;
        case 11: monthString = "November";
                 break;
        case 12: monthString = "December";
                 break;
        default: monthString = "Invalid month";
                 break;
    }
    System.out.println(monthString);
}
}

Taken from: Oracle's Java Tutorial.

score 1 · Answer 4 · answered Jul 01 '14 at 15:24

There may be better options depending on where Random is coming from and in what context this is happening, but one option you have to do basically the same thing with a different syntax is:

switch(Random) {
    case 1:
    case 2:
    case 5:
    case 7:
    case 12:
    case 14:
    case 16:
    case 18:
        label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_BLUE));
        break;
    case 3:
    case 4:
    case 6:
    case 8:
    case 9:
    case 10:
    case 11:
    case 13:
    case 15:
    case 17:
        label_wheelNumber.setForeground(SWTResourceManager.getColor(SWT.COLOR_GREEN));
        break;
}

That is only marginally better than what you have but I think it is at least a little easier to look at.

peter.petrov · Answer 5 · 2014-07-01T16:23:58.477

1

If you use a map you can turn this into a one-liner (eliminate the if/switch completely).

HashMap<Integer, Color> mp = new HashMap<Integer, Color>();

mp.put(1, SWT.COLOR_BLUE);
mp.put(2, SWT.COLOR_BLUE);
...
mp.put(18, SWT.COLOR_BLUE);

mp.put(3, SWT.COLOR_GREEN);
mp.put(4, SWT.COLOR_GREEN);
...
mp.put(17, SWT.COLOR_GREEN);


...

label_number.setForeground(SWTResourceManager.getColor(mp.get(Random)));

Also, name your Random variable differently as it clashes with a class name from the Java API.

edited Jul 01 '14 at 16:23

answered Jul 01 '14 at 15:25

peter.petrov

38,363
16
94
159

@LuiggiMendoza I am just sketching an idea here. Good that yours compiles ;) – peter.petrov Jul 01 '14 at 15:35
For a user that has more than 10k rep, you should provide code that compiles or explain that this is an idea. – Luiggi Mendoza Jul 01 '14 at 15:40
@LuiggiMendoza Take it easy. – peter.petrov Jul 01 '14 at 15:46
The only thing that hurt me here is the rawtype HashMap, the fact that compile or not.. well, it's just an example. – Marco Acierno Jul 01 '14 at 15:53

score 1 · Answer 6 · answered Jul 01 '14 at 15:28

1

Use a lookup list:

int[] ALLOW_BLUE = {1,2,5,7,12,14,16,18};
int[] ALLOW_GREEN = {3,4,6,8,9,10,11,13,15,17};

     if(Arrays.asList(ALLOW_BLUE).contains(random){
        label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_BLUE));
     }
else if(Arrays.asList(ALLOW_GREEN).contains(random){
        label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_GREEN));
     }

answered Jul 01 '14 at 15:28

Narmer

1,414
7
16

Since the values in the array are sorted, you may take advantage of this and use binary search provided by Java in `Arrays#binarySearch` so you won't have the overhead of converting the array into a List and perform a linear search. – Luiggi Mendoza Jul 01 '14 at 15:38
Also, the variable could be a list it self, there's no need to create a new one each time the if is evaluated. – OscarRyz Jul 01 '14 at 15:54
You are both right, I could have used a [double brace initialization](http://stackoverflow.com/q/924285/3735079) and a binary search on the `List`. But here it would save only some milliseconds... Indeed with those devices it's the best solution. – Narmer Jul 01 '14 at 16:01

Suresh Atta · Answer 7 · 2014-07-01T15:35:28.813

1

Switch case will do but another way you can do is

int[] blueArray ={1,2,5,7,12,14,16,18};

if(Utils.arrayContain(blueArray,Random)){//your  util method
       label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_BLUE));
}elseif(){
}

Take your possible values in an array and check Random is there in that value or not.

edited Jul 01 '14 at 15:35

answered Jul 01 '14 at 15:28

Suresh Atta

120,458
37
198
307

Roberto Linares · Answer 8 · 2014-07-01T15:35:10.340

A more elegant way would be to place each group of numbers in an ArrayList and then check if the Random number is contained in the array with the contains() method:

if(Arrays.asList(new Integer[]{1, 2, 5, 7, 12, 14, 16, 18}).contains(Random )) {
    label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_BLUE));
} else if (Arrays.asList(new Integer[]{3, 4, 6, 8, 9, 10, 11, 13, 15, 17}).contains(Random)) {
    label_wheelNumber.setForeground(SWTResourceManager.getColor(SWT.COLOR_GREEN));
}

For performance concerns you can also declare and store these arrays separately instead of creating them on the fly everytime you want to make the comparison.

OscarRyz · Answer 9 · 2014-07-01T16:14:01.847

For situations where you can't use a switch but you still have some complicated condition to evaluate you can use an "explanatory variable"

// before
if ( a && b || c != d && e > f && g < h && h == 1 ) { 
   doSomething();
} else if ( i && j || etc ) {
   doSomethingElse();
}

//after with a variable
boolean shouldDoSomething = a && b || c != d && e > f && g < h && h == 1;
if ( shuoldDoSomething ) { 
   doSomething();
} else if ( i && j || etc ) {
   doSomethingElse();
}

Or create a method that evaluates the condition:

// after with a method 
if ( shouldDoSomething(a,b,c,d,e,f,g,h) ) { 
   doSomething();
} else if ( shouldDoSomethingElse(i, j, etc ))  {
   doSomethingElse();
}
...
private boolean shouldDoSomething( boolean a,boolean b,int c,int d,int e,int f,int g,int h) {
    return a && b || c != d && e > f && g < h && h == 1;
}
private boolean shouldSoSomethingElse(boolean i, boolean j, boolean etc ) {
    return i && j || etc;
}

The objective is to simplify your code allowing you to understand it better and making it easier to modify a less error prone. If using a variable or creating a method is more confusing then continue with the simple evaluation. You can also combine the three:

//
boolean shouldDoX = a || b; 
if  ( shouldDoX || e != d && inRange(f, g, h ) ) {
  doSomething();
}

Again the objective is to make it easier to maintain.

In this example the variables are a, b, c but in real code you should use a short meaningful variable name

if ( inRange(random) ) { 
    label_number.setForeground(SWTResourceManager.getColor(SWT.COLOR_BLUE));
} else if ( outOfRange(random)) { 
    label_wheelNumber.setForeground(SWTResourceManager.getColor(SWT.COLOR_GREEN));
}  

...
private boolean inRange(int random ) {
   // use switch or simple return random == 1 || random == 2 etc. 
}

Also, final notes on style: do always use brace on your if's even if they are one line and keep the opening brace in the same line. In Java, variables start with lowercase and are written in camelStyle ( no underscores ). These are just style conventions and will help you to create good habits. Each language has its own conventions, learn them and use them.

Lots of If conditions in Java

9 Answers9