是否可以在 C# 中编写基于规则的迭代器结构的二维数组（对于网格中的瓦片的邻居）？

Question

I'm using C# and I used a 2d array of structs for a grid of tiles.This is not about how to find 8 neighboring tiles from a tile in the grid. I understand that in c# you can have a series of yield returns make a ienumerable. Like:

public IEnumerable<int> fakeList()
{
    yield return 1;
    yield return 2;
}

And call it with a foreach loop. Now, in my grid class want to have an easy way to access neighbours in grid.array[x,y] and modify it. But since it is a struct, I can't write an iterator like:

public IEnumerable<int> neighbours(int x, int y)
{
    if((x+1) >=0 && y >=0 && .....)//check if node above is inside grid
       yield return grid.array[x+1,y];
    //rinse and repeat 7 more times for each direction
}

Instead, every time I need the neighbors, I need to copy paste the 8if conditions and check that I'm using the correct x+direction,y+direction to find valid indices. Basically, a huge pain.

I could work around by:

1. Not using structs and making my life easier. And getting rid of possible premature optimization. BUT I'm going to running this code every frame in my game. Possibly multiple times. So I'd like to keep the structs if possible.
2. Write iterator for indices instead. Ex:

Is the 2nd approach valid? Or does it generate garbage? I don't know how yield return works in detail.

public struct GridTile
{
    public int x;
    public int z;

    public GridTile(int x, int z)
    {
        this.x = x;
        this.z = z;
    }
}




public IEnumerable<int> neighbours(int x, int y)
{
    if ((x + 1) >= 0 && y >= 0 && .....)//check if right node is inside
        yield return new Gridtile(x + 1, y);
    //rinse and repeat 7 more times for each direction
}

Answer 1

If you know the coordinates of a 2D array entry, then the neighbors can be retrieved using loops:

var twoD = new int[10,10];

var someX = 5;
var someY = 5;

List<int> neighbors = new List<int>();

for(int nx = -1; nx <= 1; nx++){
  for(int ny = -1; ny <= 1; ny++){
    int iX = someX + nX;
    int iY = someY + nY;
    if(iX > 0 && iX < twoD.GetLength(0) && iY > 0 && iY < twoD.GetLength(1))
      neighbors.Add(twoD[iX,iY]);
  }
}