我正在阅读有关LINQ的this article,并且无法理解在延迟评估方面如何执行查询。
因此,我将文章中的示例简化为此代码:
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文件中有以下内容:
可以从逻辑上解释。
当我想将前三个元素与前三个元素交错时,事情变得奇怪了:
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(来自上面提到的文章):
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;
}
}
执行这些代码后,我在txt文件中得到以下输出:
这让我感到尴尬,因为我不理解我的查询执行的顺序。
为什么以这种方式执行查询?
答案 0 :(得分:2)
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}");
让我们逐步完成代码:
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,代码负责前四行输出:
Executing query GetFirstSequence Executing query Take Executing query Skip Executing query Interleave
没有IEnumerable&lt; string&gt;已被评估过。
最后,shuffle.Dump()
遍历shuffle
,从而评估IEnumebles。
对data
进行迭代打印以下内容,因为SelectMany()
为GetSecondSequence()
中的每个元素调用了LogQuery()
和GetFirstSequence()
:
Executing query GetSecondSequence a Executing query GetSecondSequence b Executing query GetSecondSequence c
迭代shuffle
与迭代
Interleave(data.Take(3), data.Skip(3))
Interleave()
将data
上的两次迭代中的元素交错,因此也会迭代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"