Question

我有class StatsFile，如下所示。它有两个字段，用于保存两个字符串，这些字符串将添加到List<>：

public class StatsFile
{
    public string RemoteFileName;
    public string LocalFileName;
}

然后我有另一个class FileProcessor与method GetAllFiles填充List<>多个StatsFile s：

class FileProcessor
{
    public List<StatsFile> GetAllFiles()
    {
        List<StatsFile> fileList = new List<StatsFile>();

        StatsFile weekFiles = new StatsFile();

        weekFiles.RemoteFileName = "Remote\file1";
        weekFiles.LocalFileName = "Local\file1";
        fileList.Add(weekFiles);

        weekFiles.RemoteFileName = "Remote\file2";
        weekFiles.LocalFileName = "Local\file2";
        fileList.Add(weekFiles);

        weekFiles.RemoteFileName = "Remote\file3";
        weekFiles.LocalFileName = "Local\file3";
        fileList.Add(weekFiles);

        return fileList;
    }           
}

只是StatsFile的一个实例，List<>中的所有项目都具有相同的值。我可以实例化StatsFile 3次来解决问题，但还有另一种方法可以重用同一个实例吗？或者更好的方式？

Answer 1

您可以在构造期间初始化变量：

class FileProcessor
{
    public List<StatsFile> GetAllFiles()
    {
        List<StatsFile> fileList = new List<StatsFile>();

        fileList.Add(new StatsFile {
            RemoteFileName = "Remote\file1",
            LocalFileName = "Local\file1"
        });

        fileList.Add(new StatsFile {
            RemoteFileName = "Remote\file2",
            LocalFileName = "Local\file2"
        });

        fileList.Add(new StatsFile {
            RemoteFileName = "Remote\file3",
            LocalFileName = "Local\file3"
        });

        return fileList;
    }           
}

Answer 2

使用 struct 代替StatsFile类型的类。

public struct StatsFile
{
    public string RemoteFileName { get; set; }
    public string LocalFileName { get; set; }
}

class FileProcessor
{
    public List<StatsFile> GetAllFiles()
    {
        var fileList = new List<StatsFile>();
        StatsFile weekFiles;

        weekFiles.RemoteFileName = @"Remote\file1";
        weekFiles.LocalFileName = @"Local\file1";
        fileList.Add(weekFiles);

        weekFiles.RemoteFileName = @"Remote\file2";
        weekFiles.LocalFileName = @"Local\file2";
        fileList.Add(weekFiles);

        weekFiles.RemoteFileName = @"Remote\file3";
        weekFiles.LocalFileName = @"Local\file3";
        fileList.Add(weekFiles);

        return fileList;
    }           
}

以下是结果

FileProcessor fp = new FileProcessor();
var result = fp.GetAllFiles();
result.Dump();

Answer 3

除了此处给出的所有答案外，您还可以向StatsFile课程添加方法，并将其命名为AddCurrentInstance。此方法可以接收List<StatsFile>作为输入并将自我克隆添加到其中。

因此，StatsFile将采用以下方法：

public void AddCurrentInstance (List<StatsFile> listings)
{
    if (listings == null)
    {
        Console.WriteLine("Received null listings. Returning...");
        return;
    }
    listings.Add(this.clone());
}

现在，您的电话会变成以下内容：

public List<StatsFile> GetAllFiles()
{
    List<StatsFile> fileList = new List<StatsFile>();
    StatsFile weekFiles = new StatsFile();

    weekFiles.RemoteFileName = "Remote\file1";
    weekFiles.LocalFileName = "Local\file1";
    weekFiles.AddCurrentInstance(fileList);

    weekFiles.RemoteFileName = "Remote\file2";
    weekFiles.LocalFileName = "Local\file2";
    weekFiles.AddCurrentInstance(fileList);

    weekFiles.RemoteFileName = "Remote\file3";
    weekFiles.LocalFileName = "Local\file3";
    weekFiles.AddCurrentInstance(fileList);

    return fileList;
}

添加到List＆lt;＆gt;？时避免实例化类X次

3 个答案: