了解 C# 中 LINQ 的惰性求值

Question

我正在阅读这篇关于 LINQ 的文章，但无法理解查询是如何在惰性求值方面执行的。

因此，我将文章中的示例简化为以下代码：

void Main()
{
    var data =
        from f in GetFirstSequence().LogQuery("GetFirstSequence")
        from s in GetSecondSequence().LogQuery("GetSecondSequence", f)
        select $"{f} {s}";

    data.Dump(); // I use LINQPAD to output the data
}

static IEnumerable<string> GetFirstSequence()
{
    yield return "a";
    yield return "b";
    yield return "c";
}

static IEnumerable<string> GetSecondSequence()
{
    yield return "1";
    yield return "2";
}

public static class Extensions
{
    private const string path = @"C:\dist\debug.log";

    public static IEnumerable<string> LogQuery(this IEnumerable<string> sequence, string tag, string element = null)
    {
        using (var writer = File.AppendText(path)) 
        {
            writer.WriteLine($"Executing query {tag} {element}");
        }
        return sequence;
    }
}

执行此代码后，我在 debug.log 文件中得到了可以逻辑解释的输出：

执行查询 GetFirstSequence
执行查询 GetSecondSequence a
执行查询 GetSecondSequence b
执行查询 GetSecondSequence c

当我想将前三个元素与最后三个元素交错时，事情变得很奇怪，如下所示：

void Main()
{
    var data =
        from f in GetFirstSequence().LogQuery("GetFirstSequence")
        from s in GetSecondSequence().LogQuery("GetSecondSequence", f)
        select $"{f} {s}";

    var shuffle = data;
    shuffle = shuffle.Take(3).LogQuery("Take")
        .Interleave(shuffle.Skip(3).LogQuery("Skip")).LogQuery("Interleave");

    shuffle.Dump();
}

当然，我需要添加扩展方法来交错两个序列（来自上述文章）：

public static IEnumerable<string> Interleave(this IEnumerable<string> first, IEnumerable<string> second)
    {
        var firstIter = first.GetEnumerator();
        var secondIter = second.GetEnumerator();

        while (firstIter.MoveNext() && secondIter.MoveNext())
        {
            yield return firstIter.Current;
            yield return secondIter.Current;
        }
    }

执行这些代码行后，我在 txt 文件中得到以下输出：

执行查询 GetFirstSequence
执行查询 Take
执行查询 Skip
执行查询 Interleave 执行查询
GetSecondSequence a
执行查询 GetSecondSequence a
执行查询 GetSecondSequence b
执行查询 GetSecondSequence c
执行查询 GetSecondSequence b

这让我很尴尬，因为我不明白查询的执行顺序。

为什么查询会这样执行？

Answer 1

var data =
    from f in GetFirstSequence().LogQuery("GetFirstSequence")
    from s in GetSecondSequence().LogQuery("GetSecondSequence", f)
    select $"{f} {s}";

只是另一种写作方式

var data = GetFirstSequence()
    .LogQuery("GetFirstSequence")
    .SelectMany(f => GetSecondSequence().LogQuery("GetSecondSequence", f), (f, s) => $"{f} {s}");

让我们逐步浏览一下代码：

var data = GetFirstSequence() // returns an IEnumerable<string> without evaluating it
    .LogQuery("GetFirstSequence") // writes "GetFirstSequence" and returns the IEnumerable<string> from its this-parameter without evaluating it
    .SelectMany(f => GetSecondSequence().LogQuery("GetSecondSequence", f), (f, s) => $"{f} {s}"); // returns an IEnumerable<string> without evaluating it

var shuffle = data;
shuffle = shuffle
    .Take(3) // returns an IEnumerable<string> without evaluating it
    .LogQuery("Take") // writes "Take" and returns the IEnumerable<string> from its this-parameter without evaluating it
    .Interleave(
        shuffle
            .Skip(3) // returns an IEnumerable<string> without evaluating it
            .LogQuery("Skip") // writes "Skip" and returns the IEnumerable<string> from its this-parameter without evaluating it
    ) // returns an IEnumerable<string> without evaluating it
    .LogQuery("Interleave"); // writes "Interleave" and returns the IEnumerable<string> from its this-parameter without evaluating it

到目前为止的代码负责输出的前四行：

执行查询 GetFirstSequence
执行查询 Take
执行查询 跳过
执行查询交错

尚未评估任何 IEnumerable<string>。

最后，shuffle.Dump()迭代shuffle并评估 IEnumerables。

迭代会data打印以下内容，因为对 中的每个元素SelectMany()调用GetSecondSequence()和：LogQuery()GetFirstSequence()

执行查询 GetSecondSequence a
执行查询 GetSecondSequence b
执行查询 GetSecondSequence c

迭代shuffle与迭代相同

Interleave(data.Take(3), data.Skip(3))

Interleave()交错两次迭代的元素data，因此也交错由迭代它们引起的输出。

firstIter.MoveNext();
// writes "Executing query GetSecondSequence a"
secondIter.MoveNext();
// writes "Executing query GetSecondSequence a"
// skips "a 1" from second sequence
// skips "a 2" from second sequence
// writes "Executing query GetSecondSequence b"
// skips "b 1" from second sequence
yield return firstIter.Current; // "a 1"
yield return secondIter.Current; // "b 2"
firstIter.MoveNext();
secondIter.MoveNext();
// writes "Executing query GetSecondSequence c"
yield return firstIter.Current; // "a 2"
yield return secondIter.Current; // "c 1"
firstIter.MoveNext();
// writes "Executing query GetSecondSequence b"
secondIter.MoveNext();
yield return firstIter.Current; // "b 1"
yield return secondIter.Current; // "c 2"