如何将扩展方法作为参数传递给类的构造函数,并让该类中的方法将该扩展用作扩展?将方法扩展作为参数传递
例如:这是含有一个IEnumerable扩展方法的文件:
namespace System.Collections.Generic
{
public static partial class IEnumerableMethodExtensions
{
/// <summary>
/// Used by IsImageFile to determine if a file is a graphic type
/// we care about. (Not an Extension Method, just a helper method)
/// </summary>
public static string[] GraphicFileExtensions = new string[] { ".png", ".bmp", ".gif", ".jpg", ".jpeg" };
/// <summary>
/// Method Extension - specifies that FileInfo IEnumerable should only
/// return files whose extension matches one in GraphicFileExtensions[].
/// </summary>
/// <param name="files"></param>
/// <returns></returns>
public static IEnumerable<FileInfo> IsImageFile(this IEnumerable<FileInfo> files)
{
foreach (FileInfo file in files)
{
string ext = file.Extension.ToLower();
if (GraphicFileExtensions.Contains(ext))
yield return file;
}
}
}
}
我希望能够通过IsImageFile()作为参数传递给一个构造到该对象,所以,在这类方法可以使用IsImageFile作为方法的扩展:
public class MainFileInfoSource
{
public MainFileInfoSource(List<DirectoryInfo> Directories,
ENUMERABLE_METHOD_EXTENSION_FILE_FILTER TheMethodExtension)
{
_myFilterMethodExtension = TheMethodExtension;
_directories = Directories;
initializeFileInfoList();
}
...
/// <summary>
/// Initializes the Files list.
/// </summary>
private void initializeFileInfoList()
{
...
for (int i = 0; i < _directories.Count; i++)
{
iEnumerableFileInfo = new[] { _directories[i] }.Traverse(dir =>
getDirectoryInfosWithoutThrowing(dir)).SelectMany(dir =>
getFileInfosWithoutThrowing(dir)._myFilterMethodExtension());
http://stackoverflow.com/questions/1016033/extension-methods-defined-on-value-types-cannot-be-used-to-create-delegates-wh – AlexanderBrevig 2014-09-19 22:58:49
@AlexanderBrevig你能帮助我了解问题/答案的哪一部分是我的外卖?显然我不够聪明,看不到连接。 – 2014-09-19 23:00:54
你不能这样做,因为静态方法调用是在编译时编译的。解释你的实际目标是什么可能有助于找到好看的方法(比如使用服务定位器模式来实现的扩展方法)。 – 2014-09-19 23:15:15