blocks|key|1839631|text|复制枚举，在本例中使用IEnumerable扩展方法，并对其进行枚举。这会将每个内部可枚举中的每个元素的副本添加到该枚举中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1839632|foreach(var+item+in+Enumerable)
{
++++foreach(var+item2+in+item.Enumerable.ToList())
++++{
++++++++item.Add(item2)
++++}
}|code-block|syntax|javascript|1839633|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Make a copy of the enumeration, using an IEnumerable extension method in this case, and enumerate over it. This would add a copy of every element in every inner enumerable to that enumeration.

<pre><code>foreach(var item in Enumerable)
{
 foreach(var item2 in item.Enumerable.ToList())
 {
 item.Add(item2)
 }
}
</code></pre>

blocks|key|1839854|text|如上所述，但是使用了代码示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1839855|foreach(var+item+in+collection.ToArray())
++++collection.Add(new+Item...);|code-block|syntax|javascript|1839856|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

As mentioned, but with a code sample:

<pre><code>foreach(var item in collection.ToArray())
 collection.Add(new Item...);
</code></pre>

blocks|key|2053775|text|在这种情况下，您真的应该使用for()而不是foreach()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2053776|entityMap^0|E|5|M|9|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

You should really use <code>for()</code> instead of <code>foreach()</code> in this case.

blocks|key|1839808|text|您不能在枚举时更改可枚举集合，因此必须在枚举之前或之后进行更改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1839809|for循环是一个很好的替代方法，但是如果您的IEnumerable集合没有实现ICollection，那么它就不可能实现。|offset|length|style|CODE|1839810|以下任一项：|1839811|1)先复制集合。枚举复制的集合并在枚举期间更改原始集合。(@tvanfosson)|1839812|或|1839813|2)保存更改列表，并在枚举后提交更改。|1839814|entityMap^0|0|0|3|M|B|13|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]|$D|V|E|W|F|G]|$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|10|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|11|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|12|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|Q|$]]

You can't change the enumerable collection while it is being enumerated, so you will have to make your changes before or after enumerating.

The <code>for</code> loop is a nice alternative, but if your <code>IEnumerable</code> collection does not implement <code>ICollection</code>, it is not possible.

Either: 

1) Copy collection first. Enumerate the copied collection and change the original collection during the enumeration. (@tvanfosson)

or

2) Keep a list of changes and commit them after the enumeration.

blocks|key|1839678|text|以下是你如何做到这一点(快速和肮脏的解决方案。如果您确实需要这种行为，您应该重新考虑您的设计，或者覆盖所有IList<T>成员并聚合源列表)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1839679|using+System;
using+System.Collections.Generic;

namespace+ConsoleApplication3
{
++++public+class+ModifiableList<T>+:+List<T>
++++{
++++++++private+readonly+IList<T>+pendingAdditions+=+new+List<T>();
++++++++private+int+activeEnumerators+=+0;

++++++++public+ModifiableList(IEnumerable<T>+collection)+:+base(collection)
++++++++{
++++++++}

++++++++public+ModifiableList()
++++++++{
++++++++}

++++++++public+new+void+Add(T+t)
++++++++{
++++++++++++if(activeEnumerators+==+0)
++++++++++++++++base.Add(t);
++++++++++++else
++++++++++++++++pendingAdditions.Add(t);
++++++++}

++++++++public+new+IEnumerator<T>+GetEnumerator()
++++++++{
++++++++++++%2B%2BactiveEnumerators;

++++++++++++foreach(T+t+in+((IList<T>)this))
++++++++++++++++yield+return+t;

++++++++++++--activeEnumerators;

++++++++++++AddRange(pendingAdditions);
++++++++++++pendingAdditions.Clear();
++++++++}
++++}

++++class+Program
++++{
++++++++static+void+Main(string[]+args)
++++++++{
++++++++++++ModifiableList<int>+ints+=+new+ModifiableList<int>(new+int[]+{+2,+4,+6,+8+});

++++++++++++foreach(int+i+in+ints)
++++++++++++++++ints.Add(i+*+2);

++++++++++++foreach(int+i+in+ints)
++++++++++++++++Console.WriteLine(i+*+2);
++++++++}
++++}
}|code-block|syntax|javascript|1839680|entityMap^0|1H|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Here's how you can do that (quick and dirty solution. If you really need this kind of behavior, you should either reconsider your design or override all <code>IList&lt;T&gt;</code> members and aggregate the source list):

<pre><code>using System;
using System.Collections.Generic;

namespace ConsoleApplication3
{
 public class ModifiableList&lt;T&gt; : List&lt;T&gt;
 {
 private readonly IList&lt;T&gt; pendingAdditions = new List&lt;T&gt;();
 private int activeEnumerators = 0;

 public ModifiableList(IEnumerable&lt;T&gt; collection) : base(collection)
 {
 }

 public ModifiableList()
 {
 }

 public new void Add(T t)
 {
 if(activeEnumerators == 0)
 base.Add(t);
 else
 pendingAdditions.Add(t);
 }

 public new IEnumerator&lt;T&gt; GetEnumerator()
 {
 ++activeEnumerators;

 foreach(T t in ((IList&lt;T&gt;)this))
 yield return t;

 --activeEnumerators;

 AddRange(pendingAdditions);
 pendingAdditions.Clear();
 }
 }

 class Program
 {
 static void Main(string[] args)
 {
 ModifiableList&lt;int&gt; ints = new ModifiableList&lt;int&gt;(new int[] { 2, 4, 6, 8 });

 foreach(int i in ints)
 ints.Add(i * 2);

 foreach(int i in ints)
 Console.WriteLine(i * 2);
 }
 }
}
</code></pre>

blocks|key|1840039|text|LINQ在处理集合方面非常有效。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1840040|我不清楚你的类型和结构，但我会尽我最大的努力使你的例子符合我的能力。|1840041|从您的代码中可以看出，对于每一项，您都在向该项添加它自己的“Enumerable”属性中的所有内容。这很简单：|1840042|foreach+(var+item+in+Enumerable)
{
++++item+=+item.AddRange(item.Enumerable));
}|code-block|syntax|javascript|1840043|作为一个更一般的例子，假设我们想要迭代一个集合并删除某个条件为真的项。避免foreach，使用LINQ：|style|CODE|1840044|myCollection+=+myCollection.Where(item+=>+item.ShouldBeKept);|1840045|是否基于每个现有项目添加项目？没问题：|1840046|myCollection+=+myCollection.Concat(myCollection.Select(item+=>+new+Item(item.SomeProp)));|1840047|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Language_Integrated_Query^0|0|4|0|0|0|0|0|11|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|6|7|18|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|19|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|1A|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|1B|8|@$A|1C|B|1D|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|J|7|1E|8|@]|9|@]|C|$K|L]]|$1|S|3|T|5|6|7|1F|8|@]|9|@]|C|$]]|$1|U|3|V|5|J|7|1G|8|@]|9|@]|C|$K|L]]|$1|W|3|-4|5|6|7|1H|8|@]|9|@]|C|$]]]|X|$Y|$5|Z|10|11|C|$12|13]]]]

<a href="http://en.wikipedia.org/wiki/Language_Integrated_Query" rel="nofollow">LINQ</a> is very effective for juggling with collections.

Your types and structure are unclear to me, but I will try to fit your example to the best of my ability.

From your code it appears that, for each item, you are adding to that item everything from its own 'Enumerable' property. This is very simple:

<pre><code>foreach (var item in Enumerable)
{
 item = item.AddRange(item.Enumerable));
}
</code></pre>

As a more general example, let's say we want to iterate a collection and remove items where a certain condition is true. Avoiding <code>foreach</code>, using LINQ:

<pre><code>myCollection = myCollection.Where(item =&gt; item.ShouldBeKept);
</code></pre>

Add an item based on each existing item? No problem:

<pre><code>myCollection = myCollection.Concat(myCollection.Select(item =&gt; new Item(item.SomeProp)));
</code></pre>

blocks|key|2054176|text|从性能的角度来看，最好的方法可能是使用一个或两个数组。将列表复制到数组，对该数组执行操作，然后从该数组构建新的列表。访问数组元素比访问列表项更快，并且List<T>和T[]之间的转换可以使用快速的“批量复制”操作，从而避免了与访问单个项相关的开销。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2054177|例如，假设您有一个List<string>，并希望列表中以T开头的每个字符串后面都跟着一个项目"Boo"，而每个以"U“开头的字符串则被完全删除。最优的方法可能类似于：|2054178|int+srcPtr,destPtr;
string[]+arr;

srcPtr+=+theList.Count;
arr+=+new+string[srcPtr*2];
theList.CopyTo(arr,+theList.Count);+//+Copy+into+second+half+of+the+array
destPtr+=+0;
for+(;+srcPtr+<+arr.Length;+srcPtr%2B%2B)
{
++string+st+=+arr[srcPtr];
++char+ch+=+(st+??+"!")[0];+//+Get+first+character+of+string,+or+"!"+if+empty
++if+(ch+!=+'U')
++++arr[destPtr%2B%2B]+=+st;
++if+(ch+==+'T')
++++arr[destPtr%2B%2B]+=+"Boo";
}
if+(destPtr+>+arr.Length/2)+//+More+than+half+of+dest.+array+is+used
{
++theList+=+new+List<String>(arr);+//+Adds+extra+elements
++if+(destPtr+!=+arr.Length)
++++theList.RemoveRange(destPtr,+arr.Length-destPtr);+//+Chop+to+proper+length
}
else
{
++Array.Resize(ref+arr,+destPtr);
++theList+=+new+List<String>(arr);+//+Adds+extra+elements
}|code-block|syntax|javascript|2054179|如果List<T>提供一个从数组的一部分构造列表的方法，那会很有帮助，但我不知道有什么有效的方法可以做到这一点。尽管如此，数组上的操作还是相当快的。值得注意的是，在列表中添加和删除项并不需要“推入”其他项；每个项都直接写入数组中的相应位置。|2054180|entityMap^0|23|7|2B|3|0|9|C|T|1|0|0|2|7|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|P|$]]

The best approach from a performance perspective is probably to use a one or two arrays. Copy the list to an array, do operations on the array, and then build a new list from the array. Accessing an array element is faster than accessing a list item, and conversions between a <code>List&lt;T&gt;</code> and a <code>T[]</code> can use a fast "bulk copy" operation which avoids the overhead associated accessing individual items.

For example, suppose you have a <code>List&lt;string&gt;</code> and wish to have every string in the list which starts with <code>T</code> be followed by an item "Boo", while every string that starts with "U" is dropped entirely. An optimal approach would probably be something like:

<pre><code>int srcPtr,destPtr;
string[] arr;

srcPtr = theList.Count;
arr = new string[srcPtr*2];
theList.CopyTo(arr, theList.Count); // Copy into second half of the array
destPtr = 0;
for (; srcPtr &lt; arr.Length; srcPtr++)
{
 string st = arr[srcPtr];
 char ch = (st ?? "!")[0]; // Get first character of string, or "!" if empty
 if (ch != 'U')
 arr[destPtr++] = st;
 if (ch == 'T')
 arr[destPtr++] = "Boo";
}
if (destPtr &gt; arr.Length/2) // More than half of dest. array is used
{
 theList = new List&lt;String&gt;(arr); // Adds extra elements
 if (destPtr != arr.Length)
 theList.RemoveRange(destPtr, arr.Length-destPtr); // Chop to proper length
}
else
{
 Array.Resize(ref arr, destPtr);
 theList = new List&lt;String&gt;(arr); // Adds extra elements
}
</code></pre>

It would have been helpful if <code>List&lt;T&gt;</code> provided a method to construct a list from a portion of an array, but I'm unaware of any efficient method for doing so. Still, operations on arrays are pretty fast. Of note is the fact that adding and removing items from the list does not require "pushing" around other items; each item gets written directly to its appropriate spot in the array.

blocks|key|2054269|text|为了补充Timo的答案，LINQ也可以这样使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2054270|items+=+items.Select(i+=>+{

+++++...
+++++//perform+some+logic+adding+/+updating.

+++++return+i+/+return+new+Item();
+++++...

+++++//To+remove+an+item+simply+have+logic+to+return+null.

+++++//Then+attach+the+Where+to+filter+out+nulls

+++++return+null;
+++++...


}).Where(i+=>+i+!=+null);|code-block|syntax|javascript|2054271|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

To add to Timo's answer LINQ can be used like this as well:

<pre><code>items = items.Select(i =&gt; {

 ...
 //perform some logic adding / updating.

 return i / return new Item();
 ...

 //To remove an item simply have logic to return null.

 //Then attach the Where to filter out nulls

 return null;
 ...


}).Where(i =&gt; i != null);
</code></pre>

blocks|key|2054336|text|我已经写了一个简单的步骤，但由于这个步骤的性能将会下降|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2054337|下面是我的代码片段：|2054338|for+(int+tempReg+=+0;+tempReg+<+reg.Matches(lines).Count;+tempReg%2B%2B)
++++++++++++++++++++++++++++{
++++++++++++++++++++++++++++++++foreach+(Match+match+in+reg.Matches(lines))
++++++++++++++++++++++++++++++++{
++++++++++++++++++++++++++++++++++++var+aStringBuilder+=+new+StringBuilder(lines);
++++++++++++++++++++++++++++++++++++aStringBuilder.Insert(startIndex,+match.ToString().Replace(",",+"+");
++++++++++++++++++++++++++++++++++++lines[k]+=+aStringBuilder.ToString();
++++++++++++++++++++++++++++++++++++tempReg+=+0;
++++++++++++++++++++++++++++++++++++break;
++++++++++++++++++++++++++++++++}
++++++++++++++++++++++++++++}|code-block|syntax|javascript|2054339|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I have written one easy step, but because of this performance will be degraded

Here is my code snippet:-

<pre><code>for (int tempReg = 0; tempReg &lt; reg.Matches(lines).Count; tempReg++)
 {
 foreach (Match match in reg.Matches(lines))
 {
 var aStringBuilder = new StringBuilder(lines);
 aStringBuilder.Insert(startIndex, match.ToString().Replace(",", " ");
 lines[k] = aStringBuilder.ToString();
 tempReg = 0;
 break;
 }
 }
</code></pre>

A new feature in C# / .NET 4.0 is that you can change your enumerable in a <code>foreach</code> without getting the exception. See Paul Jackson's blog entry <a href="https://web.archive.org/web/20160308065012/www.lovethedot.net/2009/03/interesting-side-effect-of-concurrency.html" rel="noreferrer">An Interesting Side-Effect of Concurrency: Removing Items from a Collection While Enumerating</a> for information on this change.
What is the best way to do the following?
<pre><code>foreach(var item in Enumerable)
{
 foreach(var item2 in item.Enumerable)
 {
 item.Add(new item2)
 }
}
</code></pre>
Usually I use an <code>IList</code> as a cache/buffer until the end of the <code>foreach</code>, but is there better way?

What is the best way to modify a list in a 'foreach' loop?

C# / .NET 4.0中的一个新特性是，您可以在foreach中更改可枚举值，而不会出现异常。有关此更改的信息，请参阅Paul Jackson的博客条目。执行以下操作的最佳方法是什么？foreach(var item in Enumerable){    foreach(var item2 in item.Enumerable)    {        item.Add(new item2) 

问在“foreach”循环中修改列表的最好方法是什么？
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回答 9

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问在“foreach”循环中修改列表的最好方法是什么？EN

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问在“foreach”循环中修改列表的最好方法是什么？
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