blocks|key|4089104|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4089105|我曾想过，当使用相同的符号和ext调用values.Add(new+Quotes(symb,+ext));时，会返回'false‘，并且不会添加元素。|blockquote|offset|length|style|CODE|4089106|4089107|4089108|事实并非如此。|4089109|HashSet将使用GetHashCode和Equals来确定对象的相等性。现在，因为你没有覆盖Quotes中的这些方法，所以将使用默认的System.Object的引用相等。每次你添加一个新的Quote，它都是一个唯一的对象实例，所以HashSet会把它看作一个唯一的对象。|4089110|如果您覆盖Object.Equals和Object.GetHashCode，它将按照您的预期工作。|4089111|entityMap^0|0|J|Y|0|0|0|0|A|B|M|6|1C|6|1X|D|0|5|D|J|I|0^^$0|@$1|2|3|-4|4|5|6|R|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|S|7|@$D|T|E|U|F|G]]|8|@]|9|$]]|$1|H|3|-4|4|5|6|V|7|@]|8|@]|9|$]]|$1|I|3|-4|4|5|6|W|7|@]|8|@]|9|$]]|$1|J|3|K|4|5|6|X|7|@]|8|@]|9|$]]|$1|L|3|M|4|5|6|Y|7|@$D|Z|E|10|F|G]|$D|11|E|12|F|G]|$D|13|E|14|F|G]|$D|15|E|16|F|G]]|8|@]|9|$]]|$1|N|3|O|4|5|6|17|7|@$D|18|E|19|F|G]|$D|1A|E|1B|F|G]]|8|@]|9|$]]|$1|P|3|-4|4|5|6|1C|7|@]|8|@]|9|$]]]|Q|$]]

<blockquote>
 I had thought that when calling <code>values.Add(new Quotes(symb, ext));</code> with the same symb and ext, 'false' would be returned and the element would not be added.
</blockquote>

This is not the case.

HashSet will use <code>GetHashCode</code> and <code>Equals</code> to determine equality of your objects. Right now, since you're not overriding these methods in <code>Quotes</code>, the default <code>System.Object</code>'s reference equality will be used. Each time you add a new Quote, it's a unique object instance, so the HashSet sees it as a unique object.

If you override <code>Object.Equals</code> and <code>Object.GetHashCode</code>, it will work as you expect.

blocks|key|48523|text|HashSets首先根据GetHashCode计算出的散列比较条目。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|48524|默认实现根据对象本身返回一个哈希码(每个实例不同)。|48525|只有当散列相同时(对于基于实例的散列来说，这是非常不可能的)，Equals方法才会被调用并用于明确地比较两个对象。|48526|你必须选择：|48527|48528|将Quotes更改为struct|unordered-list-item|48529|Override+Quotes，并在Quotes|48530|中使用Equals|48531|示例：|48532|+public+override+int+GetHashCode()
+{
++++return+(this.symbol+==+null+?+0+:+this.symbol.GetHashCode())
+++++++%5E+(this.extension+==+null+?+0+:+this.extension.GetHashCode());
+}
+public+override+bool+Equals(object+obj)
+{
++++if+(Object.ReferenceEquals(this,+obj))
++++++return+true;

++++Quotes+other+=+obj+as+Quotes;
++++if+(Object.ReferenceEquals(other,+null))
++++++return+false;

++++return+String.Equals(obj.symbol,+this.symbol)
++++++++&&+String.Equals(obj.extension,+this.extension);
+}|code-block|syntax|javascript|48533|entityMap^0|C|B|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|17|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|18|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|19|8|@]|D|@]|E|$]]|$1|P|3|Q|5|O|7|1A|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|1B|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1C|8|@]|D|@]|E|$]]|$1|V|3|W|5|X|7|1D|8|@]|D|@]|E|$Y|Z]]|$1|10|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|11|$]]

HashSets first compare entries based on their hash which is calculated by <code>GetHashCode</code>. 
The default implementation returns a hashcode based on the object itself (differs between each instance).

Only if the hashes are the same (very improbable for hashes based on instances), the Equals method is called and used to definitely compare two objects.

You have to options:

<ul>
<li>Change Quotes to a struct</li>
<li>Override GetHashCode and Equals in Quotes</li>
</ul>

Example:

<pre><code> public override int GetHashCode()
 {
 return (this.symbol == null ? 0 : this.symbol.GetHashCode())
 ^ (this.extension == null ? 0 : this.extension.GetHashCode());
 }
 public override bool Equals(object obj)
 {
 if (Object.ReferenceEquals(this, obj))
 return true;

 Quotes other = obj as Quotes;
 if (Object.ReferenceEquals(other, null))
 return false;

 return String.Equals(obj.symbol, this.symbol)
 &amp;&amp; String.Equals(obj.extension, this.extension);
 }
</code></pre>

blocks|key|256931|text|Quotes+q+=+new+Quotes()+{+symbol+=+"GE",+extension+=+"GElec"+};
values.Add(q);
values.Add(q);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|256932|。。两次添加同一个实例，第二次将返回false。|unstyled|256933|values.Add(new+Quotes()+{+symbol+=+"GE",+extension+=+"GElec"+});
values.Add(new+Quotes()+{+symbol+=+"GE",+extension+=+"GElec"+});|256934|。。添加两个不同的实例，它们恰好具有相同的公共字段值。|256935|正如elswhere所指出的，重写Equals和GetHashCode将纠正这一点：|256936|public+class+Quotes+{+
++++public+string+symbol;+
++++public+string+extension;

++++public+override+bool+Equals(object+obj)+{
++++++++if+(!(obj+is+Quotes))+{+return+false;+}
++++++++return+(this.symbol+==+((Quotes)obj).symbol)+&&+
+++++++++++++++(this.extension+==+((Quotes)obj).extension);
++++}

++++public+override+int+GetHashCode()+{
++++++++return+(this.symbol.GetHashCode())+%5E+(this.extension.GetHashCode());
++++}
}+|256937|如果您对代码进行单步调试，您会发现values.Add同时调用Quotes.Equals和Quotes.GetHashCode。|256938|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|Z|8|@]|9|@]|A|$]]]|R|$]]

<pre><code>Quotes q = new Quotes() { symbol = "GE", extension = "GElec" };
values.Add(q);
values.Add(q);
</code></pre>

.. is adding the same instance twice, and will return false the second time.

<pre><code>values.Add(new Quotes() { symbol = "GE", extension = "GElec" });
values.Add(new Quotes() { symbol = "GE", extension = "GElec" });
</code></pre>

.. is adding two different instances that happen to have the same values for public fields.

As noted elswhere, overriding Equals and GetHashCode will correct this:

<pre><code>public class Quotes { 
 public string symbol; 
 public string extension;

 public override bool Equals(object obj) {
 if (!(obj is Quotes)) { return false; }
 return (this.symbol == ((Quotes)obj).symbol) &amp;&amp; 
 (this.extension == ((Quotes)obj).extension);
 }

 public override int GetHashCode() {
 return (this.symbol.GetHashCode()) ^ (this.extension.GetHashCode());
 }
} 
</code></pre>

If you step-debug your code, you will find that values.Add calls both Quotes.Equals and Quotes.GetHashCode.

blocks|key|1273736|text|我猜您正在创建一个具有相同值的新Quotes。在这种情况下，它们是不相等的。如果它们应该被视为相等，则重写Equals和GetHashCode方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1273737|public+class+Quotes{+
++++public+string+symbol;+
++++public+string+extension

++++public+override+bool+Equals(object+obj)
++++{
++++++++Quotes+q+=+obj+as+Quotes;
++++++++return+q+!=+null+&&+q.symbol+==+this.symbol+&&+q.extension+==+this.Extension;
++++}

++++public+override+int+GetHashCode()
++++{
++++++++return+this.symbol.GetHashCode()+%5E+this.extension.GetHashCode();
++++}
}|code-block|syntax|javascript|1273738|entityMap^0|G|6|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|$]]

I'm guessing that you are creating a new <code>Quotes</code> with the same values. In this case they are not equal. If they should be considered equal, override the Equals and GetHashCode methods.

<pre><code>public class Quotes{ 
 public string symbol; 
 public string extension

 public override bool Equals(object obj)
 {
 Quotes q = obj as Quotes;
 return q != null &amp;&amp; q.symbol == this.symbol &amp;&amp; q.extension == this.Extension;
 }

 public override int GetHashCode()
 {
 return this.symbol.GetHashCode() ^ this.extension.GetHashCode();
 }
}
</code></pre>

blocks|key|4089238|text|我知道这有点晚了，但我遇到了同样的问题，在实现选择的答案时发现了不可接受的性能影响，特别是当你有很多记录的时候。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4089239|我发现使用Hashset和Tuple将其转换为两步过程，最后通过Select进行转换，速度要快得多。|4089240|public+class+Quotes{+
++++public+string+symbol;+
++++public+string+extension
}

var+values+=+new+HashSet<Tuple<string,string>>();

values.Add(new+Tuple<string,string>("A","=n"));
values.Add(new+Tuple<string,string>("A","=n"));

//+values.Count()+==+1

values.Select+(v+=>+new+Quotes{+symbol+=+v.Item1,+extension+=+v.Item2+});|code-block|syntax|javascript|4089241|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 know this is kinda late, but I ran into the same issue and found an unacceptable performance hit while implementing the selected answer especially when you have a lot of records.

I found it much faster to turn this into a two step process using Hashset and Tuple and finally transforming via a Select.

<pre><code>public class Quotes{ 
 public string symbol; 
 public string extension
}

var values = new HashSet&lt;Tuple&lt;string,string&gt;&gt;();

values.Add(new Tuple&lt;string,string&gt;("A","=n"));
values.Add(new Tuple&lt;string,string&gt;("A","=n"));

// values.Count() == 1

values.Select (v =&gt; new Quotes{ symbol = v.Item1, extension = v.Item2 });
</code></pre>

blocks|key|1276956|text|只是想修复Kendall答案中的一些东西(由于某种奇怪的原因不能发表评论)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1276957|return+this.symbol.GetHashCode()+%5E+this.extension.GetHashCode();|code-block|syntax|javascript|1276958|请注意，xor函数是一种非常容易发生冲突的组合两个散列的方法，特别是当它们都是相同类型时(因为symbol+==扩展名将散列为0的每个对象)。即使它们不是同一类型的，或者彼此不太可能相等，这也是不好的做法，习惯了它可能会在不同的设备中导致问题。|1276959|相反，将一个散列与一个小素数相乘，然后添加第二个散列，例如：|1276960|return+3+*+this.symbol.GetHashCode()+%2B+this.extension.GetHashCode();|1276961|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Just wanted to fix something in Kendall's answer (can't comment for some strange reason).

<pre><code>return this.symbol.GetHashCode() ^ this.extension.GetHashCode();
</code></pre>

Note that the xor function is an exceptionally collision-prone way of combining two hashes, especially when they are both of the same type (since every object where symbol == extension will hash into 0). Even when they are not of the same type or are unlikely to be equal to each other, this is bad practice, and getting used to it might cause problems in different appliances.

Instead, multiply one hash with a small prime number, and add the second, e.g:

<pre><code>return 3 * this.symbol.GetHashCode() + this.extension.GetHashCode();
</code></pre>

blocks|key|48710|text|有人建议我重写Equals()和GetHashCode()不是一个好的做法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|48711|类是引用类型，结构是值类型。更改为结构将允许通过值完成相等比较，从而将相同的符号/扩展名呈现为equal。|48712|public+struct+Quotes+{+
++++public+string+symbol;+
++++public+string+extension;
}

public+static+void+Main()
{
++++var+hashSet+=+new+HashSet<Quotes>();

++++hashSet.Add(new+Quotes+{+symbol+=+"aaa",+extension+=+"bbb"+});
++++hashSet.Add(new+Quotes+{+symbol+=+"aaa",+extension+=+"bbb"+});

++++Console.WriteLine(hashSet.Count);
}|code-block|syntax|javascript|48713|输出为1。|48714|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I have been advised that overriding Equals() and GetHashCode() is not a good practice.

Classes are reference types and structs are value types. Changing to a struct will allow the equality comparison to be done by value, thus rendering identical symbol / extension as equal.

<pre><code>public struct Quotes { 
 public string symbol; 
 public string extension;
}

public static void Main()
{
 var hashSet = new HashSet&lt;Quotes&gt;();

 hashSet.Add(new Quotes { symbol = "aaa", extension = "bbb" });
 hashSet.Add(new Quotes { symbol = "aaa", extension = "bbb" });

 Console.WriteLine(hashSet.Count);
}
</code></pre>

The output is 1.

This kind of seems like a noob question, but I could not find an answer for this question specifically.

I have this class:

<pre><code>public class Quotes{ 
 public string symbol; 
 public string extension
}
</code></pre>

And am using this:

<pre><code>HashSet&lt;Quotes&gt; values = new HashSet&lt;Quotes&gt;();
</code></pre>

However I am able to add the same Quotes object multiple times. For example, my Quotes object may have 'symbol' equal to 'A' and 'extension' equal to '=n', and this Quotes object appears multiple times in the HashSet (viewing Hashset through debug mode). I had thought that when calling

<pre><code>values.Add(new Quotes(symb, ext));
</code></pre>

with the same symb and ext, 'false' would be returned and the element would not be added. I have a feeling it has something to do with comparing Quotes objects when the HashSet is adding a new object. Any help would be greatly appreciated!

HashSet allows duplicate item insertion - C#

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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这似乎是一个菜鸟问题，但我找不到一个具体的答案。我有这样一门课：public class Quotes{ public string symbol; public string extension}我用的是这个：HashSet<Quotes> values = new HashSet<Quotes>()...

问HashSet允许重复项目插入- C#
EN

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