blocks|key|1449632|text|~是一元的补码运算符--它翻转其操作数的位。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449633|~0+=+0xFFFFFFFF+=+-1|code-block|syntax|javascript|1449634|在二进制补码运算中，~x+==+-x-1|offset|length|style|CODE|1449635|~运算符可以在几乎任何借鉴了C语言语法的语言中找到，包括Objective-C/C%2B%2B/C#/Java/Javascript。|1449636|entityMap^0|0|0|A|A|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

~ is the unary one's complement operator -- it flips the bits of its operand.

<pre><code>~0 = 0xFFFFFFFF = -1
</code></pre>

in two's complement arithmetic, <code>~x == -x-1</code>

the ~ operator can be found in pretty much any language that borrowed syntax from C, including Objective-C/C++/C#/Java/Javascript.

blocks|key|1664759|text|我是这么想的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1664760|[Flags]
public+enum+PurchaseMethod
{
++++None+=+0,
++++Cash+=+1,
++++Check+=+2,
++++CreditCard+=+4,
++++All+=+Cash+%7C+Check+%7C+CreditCard
+}|code-block|syntax|javascript|1664761|会更清楚一点。|1664762|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I'd think that:

<pre><code>[Flags]
public enum PurchaseMethod
{
 None = 0,
 Cash = 1,
 Check = 2,
 CreditCard = 4,
 All = Cash | Check | CreditCard
 }
</code></pre>

Would be a bit more clear.

blocks|key|1449586|text|public+enum+PurchaseMethod
{+++
++++All+=+~0,+//+all+bits+of+All+are+1.+the+~+operator+just+inverts+bits
++++None+=++0,
++++Cash+=++1,
++++Check+=++2,
++++CreditCard+=++4
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1449587|由于二进制表示法中的二进制表示法中所有位都是1的数字，在C#中有两个补码。参阅~0+==+-1。|unstyled|offset|length|style|CODE|1449588|entityMap^0|0|13|8|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>public enum PurchaseMethod
{ 
 All = ~0, // all bits of All are 1. the ~ operator just inverts bits
 None = 0,
 Cash = 1,
 Check = 2,
 CreditCard = 4
}
</code></pre>

Because of two complement in C#, <code>~0 == -1</code>, the number where all bits are 1 in the binary representation.

blocks|key|1664840|text|它比你的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1664841|All+=+Cash+%7C+Check+%7C+CreditCard|code-block|syntax|javascript|1664842|解决方案，因为如果您稍后添加了另一个方法，例如：|1664843|PayPal+=+8+,|1664844|您已经完成了代字号-all，但必须用另一个更改all-line。这样以后就不容易出错了。|1664845|问候|1664846|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Its better than the

<pre><code>All = Cash | Check | CreditCard
</code></pre>

solution, because if you add another method later, say:

<pre><code>PayPal = 8 ,
</code></pre>

you will be already done with the tilde-All, but have to change the all-line with the other. So its less error-prone later.

regards

blocks|key|1449826|text|只是附带说明一下，当您使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449827|All+=+Cash+%7C+Check+%7C+CreditCard|code-block|syntax|javascript|1449828|另外一个好处是，当包含所有值时，Cash+%7C+Check+%7C+CreditCard将计算结果为All，而不是另一个不等于all的值(-1)。例如，如果在UI中使用三个复选框|offset|length|style|CODE|1449829|[]+Cash
[]+Check
[]+CreditCard|1449830|并将它们的值相加，然后用户将它们全部选中，您将在结果枚举中看到All。|1449831|entityMap^0|0|0|G|P|1B|3|0|0|V|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

Just a side note, when you use

<pre><code>All = Cash | Check | CreditCard
</code></pre>

you have the added benefit that <code>Cash | Check | CreditCard</code> would evaluate to <code>All</code> and not to another value (-1) that is not equal to all while containing all values.
For example, if you use three check boxes in the UI

<pre><code>[] Cash
[] Check
[] CreditCard
</code></pre>

and sum their values, and the user selects them all, you would see <code>All</code> in the resulting enum.

blocks|key|1665000|text|对于其他发现这个问题很有启发性的人，我有一个快速的~示例与您分享。以下来自paint方法实现的代码片段(如this+Mono+documentation中的详细信息)使用了~，效果很好：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1665001|PaintCells+(clipBounds,+
++++DataGridViewPaintParts.All+&+~DataGridViewPaintParts.SelectionBackground);|code-block|syntax|javascript|1665002|如果没有~运算符，代码可能如下所示：|1665003|PaintCells+(clipBounds,+DataGridViewPaintParts.Background+
++++%7C+DataGridViewPaintParts.Border
++++%7C+DataGridViewPaintParts.ContentBackground
++++%7C+DataGridViewPaintParts.ContentForeground
++++%7C+DataGridViewPaintParts.ErrorIcon
++++%7C+DataGridViewPaintParts.Focus);|1665004|..。因为枚举看起来像这样：|1665005|public+enum+DataGridViewPaintParts
{
++++None+=+0,
++++Background+=+1,
++++Border+=+2,
++++ContentBackground+=+4,
++++ContentForeground+=+8,
++++ErrorIcon+=+16,
++++Focus+=+32,
++++SelectionBackground+=+64,
++++All+=+127+//+which+is+equal+to+Background+%7C+Border+%7C+...+%7C+Focus
}|1665006|注意到这个枚举与Sean+Bright的答案相似吗？|1665007|我认为对我来说最重要的一点是，~在枚举中的运算符与在普通代码行中的运算符相同。|1665008|entityMap|0|LINK|mutability|MUTABLE|url|http://www.java2s.com/Open-Source/CSharp/2.6.4-mono-.net-core/System.Windows.Forms/System/Windows/Forms/DataGridViewRowPrePaintEventArgs.cs.htm^0|P|1|2E|1|1H|N|0|0|0|4|1|0|0|0|0|0|F|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1H|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1J|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1K|8|@$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]]]

For others who found this question illuminating, I have a quick <code>~</code> example to share. The following snippet from the implementation of a paint method, as detailed in <a href="http://www.java2s.com/Open-Source/CSharp/2.6.4-mono-.net-core/System.Windows.Forms/System/Windows/Forms/DataGridViewRowPrePaintEventArgs.cs.htm" rel="nofollow noreferrer">this Mono documentation</a>, uses <code>~</code> to great effect:

<pre><code>PaintCells (clipBounds, 
 DataGridViewPaintParts.All &amp; ~DataGridViewPaintParts.SelectionBackground);
</code></pre>

Without the <code>~</code> operator, the code would probably look something like this:

<pre><code>PaintCells (clipBounds, DataGridViewPaintParts.Background 
 | DataGridViewPaintParts.Border
 | DataGridViewPaintParts.ContentBackground
 | DataGridViewPaintParts.ContentForeground
 | DataGridViewPaintParts.ErrorIcon
 | DataGridViewPaintParts.Focus);
</code></pre>

... because the enumeration looks like this:

<pre><code>public enum DataGridViewPaintParts
{
 None = 0,
 Background = 1,
 Border = 2,
 ContentBackground = 4,
 ContentForeground = 8,
 ErrorIcon = 16,
 Focus = 32,
 SelectionBackground = 64,
 All = 127 // which is equal to Background | Border | ... | Focus
}
</code></pre>

Notice this enum's similarity to Sean Bright's answer?

I think the most important take away for me is that <code>~</code> is the same operator in an enum as it is in a normal line of code.

blocks|key|1449680|text|这是一个补码运算符，这是我经常参考的一篇关于按位运算符的文章|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449681|http://www.blackwasp.co.uk/CSharpLogicalBitwiseOps.aspx|offset|length|1449682|msdn也在他们的枚举文章中使用了它，这篇文章演示了它的更好的用法|1449683|http://msdn.microsoft.com/en-us/library/cc138362.aspx|1449684|entityMap|0|LINK|mutability|MUTABLE|url|1^0|0|0|1J|0|0|0|0|1H|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@$D|Y|E|Z|1|10]]|A|$]]|$1|J|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|C]]|Q|$5|M|N|O|A|$P|I]]]]

It's a complement operator,
Here is an article i often refer to for bitwise operators

<a href="http://www.blackwasp.co.uk/CSharpLogicalBitwiseOps.aspx" rel="noreferrer">http://www.blackwasp.co.uk/CSharpLogicalBitwiseOps.aspx</a>

Also msdn uses it in their enums article which demonstrates it use better

<a href="http://msdn.microsoft.com/en-us/library/cc138362.aspx" rel="noreferrer">http://msdn.microsoft.com/en-us/library/cc138362.aspx</a>

blocks|key|1665093|text|我个人使用的替代方案，与@Sean+Bright的答案做同样的事情，但在我看来更好，是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1665094|[Flags]
public+enum+PurchaseMethod
{
++++None+=+0,
++++Cash+=+1,
++++Check+=+2,
++++CreditCard+=+4,
++++PayPal+=+8,
++++BitCoin+=+16,
++++All+=+Cash+%2B+Check+%2B+CreditCard+%2B+PayPal+%2B+BitCoin
}|code-block|syntax|javascript|1665095|注意这些数字的二进制性质，它们都是2的幂，使得下面的断言是正确的：(a+%2B+b+%2B+c)+==+(a+%7C+b+%7C+c)。而且，%2B看起来好多了。|offset|length|style|CODE|1665096|entityMap^0|0|0|X|Q|1R|1|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

The alternative I personally use, which does the same thing than @Sean Bright's answer but looks better to me, is this one:

<pre><code>[Flags]
public enum PurchaseMethod
{
 None = 0,
 Cash = 1,
 Check = 2,
 CreditCard = 4,
 PayPal = 8,
 BitCoin = 16,
 All = Cash + Check + CreditCard + PayPal + BitCoin
}
</code></pre>

Notice how the binary nature of those numbers, which are all powers of two, makes the following assertion true: <code>(a + b + c) == (a | b | c)</code>. And IMHO, <code>+</code> looks better.

blocks|key|1450012|text|我对~做了一些实验，发现它可能有陷阱。考虑一下LINQPad的这段代码，它显示了当所有值都被“或”在一起时，所有枚举值的行为并不像预期的那样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1450013|void+Main()
{
++++StatusFilterEnum+x+=+StatusFilterEnum.Standard+%7C+StatusFilterEnum.Saved;
++++bool+isAll+=+(x+&+StatusFilterEnum.All)+==+StatusFilterEnum.All;
++++//isAll+is+false+but+the+naive+user+would+expect+true
++++isAll.Dump();
}
[Flags]
public+enum+StatusFilterEnum+{
++++++Standard+=0,
++++++Saved+=1,+++
++++++All+=+~0+
}|code-block|syntax|javascript|1450014|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|$]]

I have done some experimenting with the ~ and find it that it could have pitfalls. Consider this snippet for LINQPad which shows that the All enum value does not behave as expected when all values are ored together.

<pre><code>void Main()
{
 StatusFilterEnum x = StatusFilterEnum.Standard | StatusFilterEnum.Saved;
 bool isAll = (x &amp; StatusFilterEnum.All) == StatusFilterEnum.All;
 //isAll is false but the naive user would expect true
 isAll.Dump();
}
[Flags]
public enum StatusFilterEnum {
 Standard =0,
 Saved =1, 
 All = ~0 
}
</code></pre>

blocks|key|1450074|text|标志枚举中的每一位都表示启用(1)或禁用(0)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1450075|~运算符用于对数字的所有位进行反转。例如：00001001b变成11110110b。|1450076|因此，~0用于创建启用所有位的值，就像用于8位枚举的11111111b一样。|1450077|我只想补充一下，对于这种类型的枚举，使用按位左移运算符可能更方便，如下所示：|1450078|[Flags]
enum+SampleEnum
{
++++None+++=+0,++++++//+0000b
++++First++=+1+<<+0,+//+0001b
++++Second+=+1+<<+1,+//+0010b
++++Third++=+1+<<+2,+//+0100b
++++Fourth+=+1+<<+3,+//+1000b
++++All++++=+~0++++++//+1111b
}|code-block|syntax|javascript|1450079|entityMap^0|F|1|L|1|0|0|1|L|9|W|9|0|3|2|Q|9|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|19|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1A|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|R|$]]

Each bit in [Flags] enum means something enabled (<code>1</code>) or disabled (<code>0</code>). 
<code>~</code> operator is used to invert all the bits of the number. Example: <code>00001001b</code> turns into <code>11110110b</code>. 
So <code>~0</code> is used to create the value where all bits are enabled, like <code>11111111b</code> for 8-bit enum.
Just want to add that for this type of enums it may be more convenient to use bitwise left shift operator, like this:
<pre><code>[Flags]
enum SampleEnum
{
 None = 0, // 0000b
 First = 1 &lt;&lt; 0, // 0001b
 Second = 1 &lt;&lt; 1, // 0010b
 Third = 1 &lt;&lt; 2, // 0100b
 Fourth = 1 &lt;&lt; 3, // 1000b
 All = ~0 // 1111b
}
</code></pre>

I'm always surprised that even after using C# for all this time now, I still manage to find things I didn't know about...

I've tried searching the internet for this, but using the "~" in a search isn't working for me so well and I didn't find anything on MSDN either (not to say it isn't there)

I saw this snippet of code recently, what does the tilde(~) mean?

<pre><code>/// &lt;summary&gt;
/// Enumerates the ways a customer may purchase goods.
/// &lt;/summary&gt;
[Flags]
public enum PurchaseMethod
{ 
 All = ~0,
 None = 0,
 Cash = 1,
 Check = 2,
 CreditCard = 4
}
</code></pre>

I was a little surprised to see it so I tried to compile it, and it worked... but I still don't know what it means/does. Any help??

What is the tilde (~) in the enum definition?

Objective-C

Java

我总是感到惊讶的是，即使在使用C#这么长时间之后，我仍然设法找到我不知道的东西……我试过在互联网上搜索这个，但在搜索中使用"~“对我来说效果不是很好，我在MSDN上也没有找到任何东西(并不是说它不在那里)我最近看到了这段代码，波浪号(~)是什么意思？/// <summary>/// Enumerates the ways a customer may purchase goods./// </su

问枚举定义中的波浪号(~)是什么？
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回答 10

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问枚举定义中的波浪号(~)是什么？
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