c# – 是否可以缓存IEnumerable延迟评估结果？

我正在研究一种查找配置文件路径的方法.这需要进行两次传递：首先查找任何现有的配置文件,然后返回并找到第一个可写路径.

虽然我的特殊情况有些过分,但这让我想到：是否可以同时进行懒惰评估并防止多次枚举？

为了说明我的意思,请考虑以下代码：

public IEnumerable<string> GetPaths()
{
    Console.WriteLine("GetPaths() Returning 'one'");
    yield return "one";
    Console.WriteLine("GetPaths() Returning 'two'");
    yield return "two";
    Console.WriteLine("GetPaths() Returning 'three'");
    yield return "three";
}

public bool IsWritable(string path) => false; // testing only

如果我跑：

var paths = GetPaths();
Console.WriteLine("Searching for existing file..");
foreach (var path in paths)
{
    if (File.Exists(path))
    {
        Console.WriteLine($"Found existing file '{path}'");
    }
}

Console.WriteLine("Searching for a writable path..");
foreach (var path in paths.Reverse()) // NOTE: paths enumarated twice
{
    if (IsWritable(path))
    {
        Console.WriteLine($"Found writable path '{path}'");
    }
}

Console.WriteLine("No paths found");

如果文件’one’存在,我们得到：

Searching for existing file..
Returning 'one'
Found existing file 'one'

但是,如果没有文件存在,我们得到：

Searching for existing file..
Returning 'one'
Returning 'two'
Returning 'three'
Searching for a writable path..
Returning 'one'
Returning 'two'
Returning 'three'
No paths found

(我们两次浪费地枚举GetPaths()的结果)

一个简单的解决方法是将第一行更改为：

var paths = GetPaths().ToList();

但是,这意味着即使文件存在,输出也将是：

Returning 'one'
Returning 'two'
Returning 'three'
Searching for existing file..
Found existing file 'one'

(如我们不必要地枚举列表的其余部分)

是否有(内置)方法来获取延迟枚举并仍然阻止多次枚举？

换句话说,’one’存在时所需的输出是：

Searching for existing file..
Returning 'one'
Found existing file 'one'

如果没有文件：

Searching for existing file..
Returning 'one'
Returning 'two'
Returning 'three'
Searching for a writable path..
No paths found

static System.IO.FileStream fs;
private static void GetFileStream()
{
    if(fs == null) fs = System.IO.File.Open(....);
    return fs;
}

public Enumarable<string> GetLines()
{
    // return an enumerator that uses the result of GetFileStream() and 
    // advances the file pointer. All Enumerables returned by this method
    // will return unique lines from the same file
}