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我有一个结构如下,

struct Location
{
    public int Row;
    public int Column;

    public Location(int row, int column)
    {
        this.Row = row;
        this.Column = column;
    }
}

我有一个功能如下,

public List<Location> getNeighboringLocations(int row, int column)
{
    int[,] array = new int[rows, columns];
    int refx = row;
    int refy = column;

    //var neighbours = from x in Enumerable.Range(refx - 1, 3)
    //                 from y in Enumerable.Range(refy - 1, 3)
    //                 where x >= 0 && y >= 0 && x < array.GetLength(0) && y < array.GetLength(1)
    //                 select new { x, y };
    var neighbours = from x in Enumerable.Range(0, array.GetLength(0)).Where(x => Math.Abs(x - refx) <= 1)
                 from y in Enumerable.Range(0, array.GetLength(1)).Where(y => Math.Abs(y - refy) <= 1)
                 select new { x, y };

    return neighbours.ToList();
}

我希望返回类型是位置列表,我该怎么做?提前致谢

4

3 回答 3

2

...

select new Location(x, y);
于 2011-04-17T07:16:04.173 回答
2
var neighbours = from x in Enumerable.Range(0, array.GetLength(0)).Where(x => Math.Abs(x - refx) <= 1)
                             from y in Enumerable.Range(0, array.GetLength(1)).Where(y => Math.Abs(y - refy) <= 1)
                             select new Location( x, y );

return neighbours.ToList();
于 2011-04-17T07:16:49.787 回答
1

而不是这样做select new { x, y }返回一个匿名类型,你应该这样做select new Location(x, y)

于 2011-04-17T07:18:32.377 回答