Finding 8 neighbours in 2d Array

Question

I have looked around for this question, there are some answers about the question, but none that i really understand/or is not suitable for me.

So my problem is to check for 8 neighbors in an 2d array containing chars, either * or O.

Code:

aliveCheck = isAlive(g,row,column-1);
if(aliveCheck){
    aliveCounter++;
}

aliveCheck = isAlive(g,row,column+1);
if(aliveCheck == 1){
    aliveCounter++;
}

aliveCheck = isAlive(g,row+1,column);
if(aliveCheck == 1){
    aliveCounter++;
}

Etc for all the 8 neighbours, this works, but I'm not happy with the solution.

isAlive() is a simple function to findout if a coordinate is * or O.

Anyone got a better solution to this problem or got any tips on how to improve it?

Thanks

You could try looping with row and column offsets and writing one block of code that adds the offets and only changing the offsets each loop. — Avery, Nov 04 '13 at 23:07
The code gets really long when there are 8 if statements after each other, and im interested in learning some new ways to aproach this problem @nhgrif — uzr, Nov 04 '13 at 23:07
just put it in a function. To me, using the loop method described in the answers makes it more confusing of what you are trying to accomplish. — clcto, Nov 04 '13 at 23:17

nhgrif · Accepted Answer · 2013-11-04T23:18:48.383

2

for(int i=-1, i<=1; ++i) {
    for(int j=-1; j<=1; ++j {
        if((i || j) && isAlive(g,row+i,column+j)) {
            aliveCounter++; } } }

This method assumes that i-1, i+1, j-1, and j+1 are all within the bounds of your array.

It should also be noted that while this approach accomplishes what you're trying to accomplish in a very few number of lines, it's far less readable. As such, this approach should be accompanied with very descriptive comments. Moreover, this approach (or any other method) would be best wrapped in an appropriately named function (such as checkNeighbors).

edited Nov 04 '13 at 23:18

answered Nov 04 '13 at 23:08

nhgrif

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I was just about to say that! OP: Watch your boundaries on the array. – Charlie Burns Nov 04 '13 at 23:09
You could say 'if(i && j && isAlive(g,row+i,column+j) { ... }' – Charlie Burns Nov 04 '13 at 23:10
Hmm, really good point @CharlieBurns Except, should be `(i || j) && isAlive(/*etc*/)` – nhgrif Nov 04 '13 at 23:10
1

No, but I understood what you meant. ;) – nhgrif Nov 04 '13 at 23:14
This also checks at `row,column` (not sure if that is acceptable) – clcto Nov 04 '13 at 23:16
1

@clcto No it doesn't. If `i` and `j` are both `0`, then `(i || j)` returns `false` and so the `&&` we're checking in the `if` statement returns `false` and `aliveCounter` is not incremented. – nhgrif Nov 04 '13 at 23:17
@nhgrif you have if `i==0 || j==0` ... oh you just changed it. – clcto Nov 04 '13 at 23:19
Or even just two loops and "aliveCounter += (i || j) && isAlive(g,row+i,column+j); That may be getting kind of ugly. – Charlie Burns Nov 04 '13 at 23:38

Finding 8 neighbours in 2d Array

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