blocks|key|3609496|text|如果两个数字之间没有差异，则它们是相等的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3609497|int+equal(int+x,+int+y){
+++return+!(x-y);
}|code-block|syntax|javascript|3609498|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|$]]

Two numbers are equal if there is no difference between them:

<pre><code>int equal(int x, int y){
 return !(x-y);
}
</code></pre>

blocks|key|1322114|text|请记住，XOR与NOT+EQUALS完全相同，XNOR与EQUALS完全相同。因此，下面的代码将完全满足您的需求：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1322115|return+!(x+%5E+y);|code-block|syntax|javascript|1322116|entityMap^0|4|3|8|A|N|4|S|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

Remember that an <code>XOR</code> is the exactly same as <code>NOT EQUALS</code> and <code>XNOR</code> is exactly the same as <code>EQUALS</code>. So, the following will give you exactly what you want:

<pre><code>return !(x ^ y);
</code></pre>

blocks|key|1322213|text|C+!操作符实际上只是!=+0的简写，所以使用它似乎非常接近欺骗:)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1322214|下面是我仅使用位操作的观点，假设32位二进制补码机器具有算术右移位(从技术上讲，在C中，右移算术是未定义的，但我在二进制补码机器上见过的每个C编译器都正确地支持这一点)：|1322215|int+t+=+(x+-+y)+%7C+(y+-+x);+//+<0+iff+x+!=+y,+0+otherwise
t+>>=+31;+//+-1+iff+x+!=+y,+0+otherwise
return+1+%2B+t;+//+0+iff+x+!=+y,+1+otherwise|code-block|syntax|javascript|1322216|也就是说，实际的编译器没有这个问题。真正的硬件实际上直接支持比较。具体细节取决于架构，但有两个基本模型：|1322217|算术运算返回的|1322218|条件码(例如，x86和ARM执行此操作)。在这种情况下，通常有一个“比较”指令，它减去两个值，不写回整数寄存器，但根据结果设置条件代码/标志。|ordered-list-item|1322219|更类似于RISC的平台通常有直接的“等于分支”和“小于分支”操作数，它们根据结果进行比较和分支。它基本上等同于C代码|1322220|1322221|如果(a+==+b)转到标签；|1322222|或|1322223|如果(a+<+b)转到标签；|1322224|all+in+one机器指令。|1322225|entityMap^0|2|1|B|4|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1B|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1D|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|R|5|S|7|1F|8|@]|D|@]|E|$]]|$1|T|3|U|5|S|7|1G|8|@]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|1I|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1J|8|@]|D|@]|E|$]]|$1|10|3|11|5|6|7|1K|8|@]|D|@]|E|$]]|$1|12|3|13|5|6|7|1L|8|@]|D|@]|E|$]]|$1|14|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|15|$]]

The C <code>!</code> operator is really just shorthand for <code>!= 0</code>, so using it seems very close to cheating :)

Here's my take just using bitwise operations, assuming a 32-bit two's complement machine with arithmetic right shifts (technically, in C arithmetic right shifts are undefined, but every C compiler I've ever seen on a two's complement machine supports this correctly):

<pre><code>int t = (x - y) | (y - x); // &lt;0 iff x != y, 0 otherwise
t &gt;&gt;= 31; // -1 iff x != y, 0 otherwise
return 1 + t; // 0 iff x != y, 1 otherwise
</code></pre>

That said, actual compilers don't have this problem. Real hardware actually has direct support for comparisons. The details depend on the architecture, but there's two basic models: 

<ol>
<li>Condition codes returned for arithmetic operations (e.g. x86 and ARM do this). In this case, there's usually a "compare" instruction which subtracts two values, doesn't write back to an integer register but sets the condition code/flags based on the result.</li>
<li>More RISC-like platforms typically have direct "branch if equal" and "branch if less than" operands that do a comparison and branch based on the result. It's basically equivalent to the C code

<pre><code>if (a == b) goto label;
</code></pre>

or

<pre><code>if (a &lt; b) goto label;
</code></pre>

all in one machine instruction.</li>
</ol>

blocks|key|5020678|text|此示例与减法相同，但对于某些体系结构如何进行寄存器比较(如ARM，我相信)更为明确。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5020679|return+!(1+%2B+~x+%2B+y);|code-block|syntax|javascript|5020680|1表示输入到ALU的进位位。一个数字x是按位补码的。取补码并加1产生该数的2的补码(x变成-x)，然后将其与另一个数相加，得到差值以确定相等。|offset|length|style|CODE|5020681|因此，如果这两个数字相等，则得到-x+%2B+x+=>+0。|5020682|(在寄存器级别上，!运算符并未完成，您只需测试条件代码或标志寄存器的“零位”，如果寄存器操作产生的结果为零，则将其设置，否则将清除。)|5020683|entityMap^0|0|0|I|1|16|1|19|2|0|G|B|0|9|1|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]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$I|12|J|13|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

This example is the same as subtraction, but is more explicit as to how some architectures do register comparison (like the ARM, I believe).

<pre><code>return !(1 + ~x + y);
</code></pre>

The 1 signifies the carry-bit input into the ALU. One number <code>x</code> is bitwise complemented. Taking the complement and adding 1 produces the two's complement of the number (<code>x</code> becomes <code>-x</code>), and then it's added to the other number to get the difference to determine equality.

So if both numbers are equal, you get <code>-x + x =&gt; 0</code>. 

(On a register level the <code>!</code> operator isn't done, and you just test the "zero bit" of the condition codes or flags register, which gets set if the register operation produces a result of zero, and is clear otherwise.)

blocks|key|3609645|text|由于XOR与(!=)相同，因此(x%5Ey)仅对相等的值返回0。我的观点如下，因为它是合理的，使用逐位运算符和工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3609646|int+notEqual(int+x,+int+y){
++++++++return+(x+%5E+y);
}|code-block|syntax|javascript|3609647|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 XOR is same as (!=), hence (x ^ y) will return 0 only for equal values.
My take is the following because it is sensible, uses bit-wise operator and working.

<pre><code>int notEqual(int x, int y){
 return (x ^ y);
}
</code></pre>

blocks|key|3609572|text|我对此的看法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3609573|int+equal(int+x,+int+y){
+++if((x+&+~y)+==+0)
+++++++return+1;
+++else
+++++++return+0;+
}|code-block|syntax|javascript|3609574|说明:如果为x+==+y，则x+&+~y的计算结果为0返回1，否则返回0作为x!=y。|offset|length|style|CODE|3609575|Edit1:+The+above+is+equivalent+to+

int+equal(int+x,+int+y){
++++return+!(x+&+~y)+;+//+returns+1+if+equal+,+0+otherwise.+
}|3609576|在某些情况下，当最高有效位变为1时，上面的代码会失败。解决方案是添加一个1。|3609577|return+!(x+&+(~y+%2B1)+);|3609578|entityMap^0|0|0|6|6|E|6|Q|1|12|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|T|$]]

My Take on this 

<pre><code>int equal(int x, int y){
 if((x &amp; ~y) == 0)
 return 1;
 else
 return 0; 
}
</code></pre>

Explanation: If <code>x == y</code>, then <code>x &amp; ~y</code> evaluates to <code>0</code> return 1, else return 0 as <code>x!=y</code>. 

<pre><code>Edit1: The above is equivalent to 

int equal(int x, int y){
 return !(x &amp; ~y) ; // returns 1 if equal , 0 otherwise. 
}
</code></pre>

The above code fails in certain cases where the Most significant bit turns to 1. The solution is to add a 1. i.e correct answer is 

<pre><code>return !(x &amp; (~y +1) );
</code></pre>

Using only bitwise operators (|, &amp;, ~, ^, >>, &lt;&lt;) and other basic operators like +, -, and !, is it possible to replace the "==" below?

<pre><code>int equal(int x, int y) {
 return x == y;
}
</code></pre>

Replacing "==" with bitwise operators

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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仅使用按位运算符(|、&、~、^、>>、<<)和其他基本运算符(如+、-和！)，是否可以替换下面的"==“？int equal(int x, int y) { return x == y;}

问用位运算符替换"==“
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回答 6

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问用位运算符替换"==“
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