c - 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

Answer 1

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).