计算二维数组中的所有数字

Question

I'm trying to write console minesweeper, but cant figure out how to iterate over 2d array, and count amount of "-1" around each of element. My array looks like

[0 0 0 0 0]
[0 0 -1 0 0]
[0 0 0 0 0]
[0 0 0 -1 -1]
[0 0 0 -1 -1]

Where "-1" that is mine. I've tried to simply iterate over it with code

for i := 0; i < row; i++ {
numb := 0
    for j := 0; j < col; j++ {
         if ary[i-1][j-1] == -1 {
             numb ++ } 
         }
         if ary[i-1][j] == 1 {
             numb ++ }
         //rest of code omitted

    }

But that resulted in huge amout of code, which is hard to read and understand. Is there more flexible way, to check all elements around current in for-iterator, so in result it would look like

[0 1  1   1 0]
[0 1 -1   1 0]
[0 1  2   2 1]
[0 0  1  -1 -1]
[0 0  1  -1 -1]

Answer 1

for i := 0; i < row; i++ {
    for j := 0; j < col; j++ {
        if ary[i][j] == -1 {
            for k := max(0, i - 1); k <= min(row - 1, i + 1); k++ {
                for l := max(0, j - 1); l <= min(col - 1, j + 1); l++ {
                    if ary[k][l] != -1 {
                        ary[k][l] ++
                    }
                }
            } 
        }
    }
}

The first two outer loops go through the array. If the observed number is -1 then it increases all surrounding cells unless there is a mine on that field. To avoid the checks if ary[k][l] is not outside of bounds the bounds are calculated before by using the trick with min / max to ensure that k and l are not outside of bounds. E.g. max(0, i-1) will always return i-1 unless i-1 is smaller than 0, then it returns 0.