如何将扩展方法作为参数传递给类的构造函数,并使该类中的方法将其用作扩展?
例如:这是一个包含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;
}
}
}
我希望能够将IsImageFile()作为参数传递给该对象的构造函数,以便该类中的方法可以使用IsImageFile作为方法扩展:
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());