blocks|key|636819|text|这个问题有点过时了，但是...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|636820|这就是运算符的工作方式：|636821|true+xor+false+=+true
true+xor+true+=+false
false+xor+true+=+true
false+xor+false+=+false|code-block|syntax|javascript|636822|这就是!=运算符处理bool类型的方式：|636823|(true+!=+false)+//+true
(true+!=+true)+//+false
(false+!=+true)+//+true
(false+!=+false)+//+false|636824|因此，如您所见，现有的%5E%5E可以替换为不存在的!=|offset|length|style|CODE|636825|entityMap^0|0|0|0|0|0|B|2|M|2|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@$O|10|P|11|Q|R]|$O|12|P|13|Q|R]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

Question is a bit outdated but...

It's how this operator should work:

<pre><code>true xor false = true
true xor true = false
false xor true = true
false xor false = false
</code></pre>

This is how != operator works with bool types:

<pre><code>(true != false) // true
(true != true) // false
(false != true) // true
(false != false) // false
</code></pre>

So as you see unexisting <code>^^</code> can be replaced with existing <code>!=</code>

blocks|key|852933|text|有一个逻辑XOR运算符：%5E|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|852934|文档：C#+Operators和%5E+Operator|852935|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.microsoft.com/en-us/dotnet/csharp/language-reference/operators/boolean-logical-operators|1|https://docs.microsoft.com/en-us/dotnet/csharp/language-reference/operators/boolean-logical-operators#logical-exclusive-or-operator-^0|C|1|0|3|C|0|G|A|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@$9|V|A|W|1|X]|$9|Y|A|Z|1|10]]|E|$]]|$1|H|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]]]

There is the logical XOR operator: <code>^</code>

Documentation: <a href="https://docs.microsoft.com/en-us/dotnet/csharp/language-reference/operators/boolean-logical-operators" rel="noreferrer">C# Operators</a> and <a href="https://docs.microsoft.com/en-us/dotnet/csharp/language-reference/operators/boolean-logical-operators#logical-exclusive-or-operator-" rel="noreferrer">^ Operator</a>

blocks|key|636874|text|根据Mark+L的要求，以下是正确的版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|636875|+Func<bool,+bool,+bool>+XOR+=+(X,Y)+=>+((!X)+&&+Y)+%7C%7C+(X+&&+(!Y));|code-block|syntax|javascript|636876|以下是真值表：|636877|+X+%7C+Y+%7C+Result
+==============
+0+%7C+0+%7C+0
+1+%7C+0+%7C+1
+0+%7C+1+%7C+1
+1+%7C+1+%7C+0|636878|参考：Exclusive+OR|636879|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/85269/marc-l|1|https://en.wikipedia.org/wiki/Exclusive_or^0|2|6|0|0|0|0|0|3|C|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

As asked by <a href="https://stackoverflow.com/users/85269/marc-l">Mark L</a>, Here is the correct version:

<pre><code> Func&lt;bool, bool, bool&gt; XOR = (X,Y) =&gt; ((!X) &amp;&amp; Y) || (X &amp;&amp; (!Y));
</code></pre>

Here is the truth table:

<pre><code> X | Y | Result
 ==============
 0 | 0 | 0
 1 | 0 | 1
 0 | 1 | 1
 1 | 1 | 0
</code></pre>

Reference:
 <a href="https://en.wikipedia.org/wiki/Exclusive_or" rel="noreferrer">Exclusive OR</a>

blocks|key|636587|text|哦，是的，确实如此。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|636588|bool+b1+=+true;
bool+b2+=+false;
bool+XOR+=+b1+%5E+b2;|code-block|syntax|javascript|636589|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|$]]

Oh yes, it does. 

<pre><code>bool b1 = true;
bool b2 = false;
bool XOR = b1 ^ b2;
</code></pre>

blocks|key|636706|text|条件xor不存在，但您可以使用逻辑1，因为xor是为布尔值定义的，并且所有条件比较的计算结果都是布尔值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|636707|所以你可以这样说：|636708|if+(+(a+==+b)+%5E+(c+==+d))
{

}|code-block|syntax|javascript|636709|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|$]]

Conditional xor doesn't exist, but you can use logical one because xor is defined for booleans, and all conditional comparisons evaluate to booleans.

So you can say something like:

<pre><code>if ( (a == b) ^ (c == d))
{

}
</code></pre>

blocks|key|853015|text|您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|853016|a+=+b+%5E+c;|code-block|syntax|javascript|853017|就像在c/c%2B%2B中|853018|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|$]]

you can use:

<pre><code>a = b ^ c;
</code></pre>

just like in c/c++

blocks|key|636941|text|没有条件(短路)+XOR这回事。条件运算符只有在只看第一个参数就能确定最终结果的情况下才有意义。XOR+(与加法)总是需要两个参数，所以没有办法在第一个参数之后省略。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|636942|如果你知道A=true，那么(A+XOR+B)+=+!B。|636943|如果你知道A=false，那么(A+XOR+B)+=+B。|636944|在这两种情况下，如果你知道A但不知道B，那么你知道的还不够多(A+XOR+B)。您必须始终了解A和B的值才能计算出答案。没有任何用例可以在没有这两个值的情况下解析XOR。|636945|请记住，XOR的定义有四种情况：|636946|false+xor+true++=+true
true++xor+false+=+true
true++xor+true++=+false
false+xor+false+=+false|code-block|syntax|javascript|636947|同样，希望从上面可以明显看出，知道第一个值永远不足以在不知道第二个值的情况下获得答案。但是，在您的问题中，您省略了第一种情况。如果你想要|636948|false+op+true++=+false+(or+DontCare)
true++op+false+=+true
true++op+true++=+false
false+op+false+=+false|636949|那么你确实可以通过一个短路的条件运算来得到它：|636950|A+&&+!B|636951|但这不是异或运算。|636952|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|15|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|L|7|17|8|@]|9|@]|A|$M|N]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|L|7|19|8|@]|9|@]|A|$M|N]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

There is no such thing as conditional (short-circuiting) XOR. Conditional operators are only meaningful when there's a way to definitively tell the final outcome from looking at only the first argument. XOR (and addition) always require two arguments, so there's no way to short-circuit after the first argument.

If you know A=true, then (A XOR B) = !B.

If you know A=false, then (A XOR B) = B.

In both cases, if you know A but not B, then you don't know enough to know (A XOR B). You must always learn the values of both A and B in order to calculate the answer. There is literally no use case where you can ever resolve the XOR without both values.

Keep in mind, XOR by definition has four cases:

<pre><code>false xor true = true
true xor false = true
true xor true = false
false xor false = false
</code></pre>

Again, hopefully it's obvious from the above that knowing the first value is never enough to get the answer without also knowing the second value. However, in your question, you omitted the first case. If you instead wanted

<pre><code>false op true = false (or DontCare)
true op false = true
true op true = false
false op false = false
</code></pre>

then you can indeed get that by a short-circuiting conditional operation:

<pre><code>A &amp;&amp; !B
</code></pre>

But that's not an XOR.

blocks|key|853182|text|这个问题已经得到了深情的回答，但我遇到了不同的情况。确实不需要条件XOR。也可以使用%5E运算符。然而，如果你只需要测试操作数的"true+%7C%7C+false“状态，那么%5E可能会带来麻烦。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|853183|void+Turn(int+left,+int+right)
{
++++if+(left+%5E+right)
++++{
++++++++//success...+turn+the+car+left+or+right...
++++}
++++else
++++{
++++++++//error...+no+turn+or+both+left+AND+right+are+set...
++++}
}|code-block|syntax|javascript|853184|在本例中，如果left设置为10+(0xa)，right设置为5+(0x5)，则进入“成功”分支。对于这个(简单但愚蠢的)例子，这将导致一个错误，因为你不应该同时向左和向右转。我从提问者那里收集到的并不是他实际上想要一个条件，而是一种简单的方法来对传递给xor的值执行true/false。|853185|宏可以做到这一点：|853186|#define+my_xor(a,+b)+(+((a)?1:0)+%5E+((b)?1:0)+)|853187|如果我说错了，请随意扇我耳光:o)|853188|在我发布了这篇文章(糟糕的Yapdog!)之后，我读到了下面jimreed的答案。而他的实际上更简单。这是可行的，我完全不知道为什么他的答案被否决了…|853189|entityMap^0|0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

This question has been affectively answered, but I came across a different situation. It's true that there is no need for a conditional XOR. It's also true that the ^ operator can be used. However, if you need to only test the "true || false" status of the operands then ^ can lead to trouble. For example:

<pre><code>void Turn(int left, int right)
{
 if (left ^ right)
 {
 //success... turn the car left or right...
 }
 else
 {
 //error... no turn or both left AND right are set...
 }
}
</code></pre>

In this example, if left is set to 10 (0xa) and right is set to 5 (0x5) the "success" branch is entered. For this (simplistic if silly) example, this would result in a bug since you shouldn't turn left AND right at the same time. What I gathered from the questioner is not that he actually wanted a conditional, but a simple way to perform the true/false on the values as passed to the xor.

A macro could do the trick:

<pre><code>#define my_xor(a, b) ( ((a)?1:0) ^ ((b)?1:0) )
</code></pre>

Feel free to slap me around if I'm off the mark :o)

I read jimreed's answer below after I posted this (bad Yapdog!) and his is actually simpler. It would work and I have absolutely no idea why his answer was voted down...

How come C# doesn't have a conditional <code>XOR</code> operator?

Example:

<pre><code>true xor false = true
true xor true = false
false xor false = false
</code></pre>

Conditional XOR?

为什么C#没有条件XOR运算符？示例：true  xor false = truetrue  xor true  = falsefalse xor false = false

问条件异或？
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