More efficient way to check neighbours in a two-dimensional array in Java

Question

Hey all, for a few of my college assignments I've found the need to check neighbouring cells in 2-dimensional arrays (grids). The solution I've used is a bit of a hack using exceptions, and I'm looking for a way to clean it up without having loads of if statements like some of my classmates. My current solution is

for ( int row = 0; row < grid.length; row++ ) {
    for ( int col = 0; col < grid.length; col++ ) {
        // this section will usually be in a function
        // checks neighbours of the current "cell"
        try {
            for ( int rowMod = -1; rowMod <= 1; rowMod++ ) {
                for ( int colMod = -1; colMod <= 1; colMod++ ) {
                    if ( someVar == grid[row+rowMod][col+colMod] ) {
                        // do something
                    }
                }
            }
        } catch ( ArrayIndexOutOfBoundsException e ) {
            // do nothing, continue
        }
        // end checking neighbours
    }
}

I shudder to think of the inefficiency using exceptions to get my code to work causes, so I'm looking for suggestions as to how I could remove the reliance on exceptions from my code without sacrificing readability if it's possible, and just how I could make this code segment generally more efficient. Thanks in advance.

Answer 1

You can try this. First decide the size of the grid Lets say its 8 X 8 & assign MIN_X = 0, MIN_Y = 0, MAX_X =7, MAX_Y =7

Your curren position is represented by thisPosX , thisPosY, then try this:

int startPosX = (thisPosX - 1 < MIN_X) ? thisPosX : thisPosX-1;
int startPosY = (thisPosY - 1 < MIN_Y) ? thisPosY : thisPosY-1;
int endPosX =   (thisPosX + 1 > MAX_X) ? thisPosX : thisPosX+1;
int endPosY =   (thisPosY + 1 > MAX_Y) ? thisPosY : thisPosY+1;


// See how many are alive
for (int rowNum=startPosX; rowNum<=endPosX; rowNum++) {
    for (int colNum=startPosY; colNum<=endPosY; colNum++) {
        // All the neighbors will be grid[rowNum][colNum]
    }
}

you can finish it in 2 loops.