blocks|key|1393186|text|一般形式是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1393187|A+=+A+operation+B
B+=+A+inverse-operation+B
A+=+A+inverse-operation+B+|code-block|syntax|javascript|1393188|但是，您必须潜在地注意溢出，而且并不是所有的操作都有一个为该操作定义的所有值定义良好的逆值。例如*和/直到A或B为0为止|1393189|xor特别令人满意，因为它是为所有int定义的，并且是它自己的逆数|1393190|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

the general form is:

<pre><code>A = A operation B
B = A inverse-operation B
A = A inverse-operation B 
</code></pre>

however you have to potentially watch out for overflows and also not all operations have an inverse that is well defined for all values that the operation is defined. e.g. * and / work until A or B is 0

xor is particularly pleasing as it is defined for all ints and is its own inverse

blocks|key|1393093|text|a+=+a+%2B+b
b+=+a+-+b+//+b+=+a
a+=+a+-+b|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1393094|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

blocks|key|1393147|text|正如manu已经提到的，XOR算法是一种流行的算法，它适用于所有整数值(包括指针，然后加上一些运气和强制转换)。为了完整起见，我想提到另一个功能较弱的加/减算法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1393148|A+=+A+%2B+B
B+=+A+-+B
A+=+A+-+B|code-block|syntax|javascript|1393149|在这里，你必须小心溢出/下溢，但除此之外，它也能正常工作。您甚至可以在浮点数/双精度数上尝试此操作，因为不允许对这些浮点数/双精度数执行XOR运算。|1393150|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|$]]

As already noted by manu, XOR algorithm is a popular one which works for all integer values (that includes pointers then, with some luck and casting). For the sake of completeness I would like to mention another less powerful algorithm with addition/subtraction:

<pre><code>A = A + B
B = A - B
A = A - B
</code></pre>

Here you have to be careful of overflows/underflows, but otherwise it works just as fine. You might even try this on floats/doubles in the case XOR isn't allowed on those.

blocks|key|1393211|text|愚蠢的问题应该得到适当的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1393212|void+sw2ap(int&+a,+int&+b)+{
++register+int+temp+=+a;+//+!
++a+=+b;
++b+=+temp;
}|code-block|syntax|javascript|1393213|register关键字的唯一好用法。|offset|length|style|CODE|1393214|entityMap^0|0|0|0|8|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]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Stupid questions deserve appropriate answers:

<pre><code>void sw2ap(int&amp; a, int&amp; b) {
 register int temp = a; // !
 a = b;
 b = temp;
}
</code></pre>

The only good use of the <code>register</code> keyword.

blocks|key|1178185|text|使用第三个变量交换两个数字如下所示，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1178186|int+temp;
int+a=10;
int+b=20;
temp+=+a;
a+=+b;
b+=+temp;
printf+("Value+of+a",+%25a);
printf+("Value+of+b",+%25b);|code-block|syntax|javascript|1178187|不使用第三个变量交换两个数字|1178188|int+a+=+10;
int+b+=+20;
a+=+a%2Bb;
b+=+a-b;
a+=+a-b;
printf+("value+of+a=",+%25a);
printf+("value+of+b=",+%25b);|1178189|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Swapping two numbers using third variable be like this,

<pre><code>int temp;
int a=10;
int b=20;
temp = a;
a = b;
b = temp;
printf ("Value of a", %a);
printf ("Value of b", %b);
</code></pre>

Swapping two numbers without using third variable

<pre><code>int a = 10;
int b = 20;
a = a+b;
b = a-b;
a = a-b;
printf ("value of a=", %a);
printf ("value of b=", %b);
</code></pre>

blocks|key|1177720|text|#include<iostream.h>
#include<conio.h>
void+main()
{
int+a,b;
clrscr();
cout<<"\n==========Vikas==========";
cout<<"\n\nEnter+the+two+no=:";
cin>>a>>b;
cout<<"\na"<<a<<"\nb"<<b;
a=a%2Bb;
b=a-b;
a=a-b;

cout<<"\n\na="<<a<<"\nb="<<b;
getch();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1177721|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include&lt;iostream.h&gt;
#include&lt;conio.h&gt;
void main()
{
int a,b;
clrscr();
cout&lt;&lt;"\n==========Vikas==========";
cout&lt;&lt;"\n\nEnter the two no=:";
cin&gt;&gt;a&gt;&gt;b;
cout&lt;&lt;"\na"&lt;&lt;a&lt;&lt;"\nb"&lt;&lt;b;
a=a+b;
b=a-b;
a=a-b;

cout&lt;&lt;"\n\na="&lt;&lt;a&lt;&lt;"\nb="&lt;&lt;b;
getch();
}
</code></pre>

blocks|key|1177892|text|由于原始解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177893|temp+=+x;+y+=+x;+x+=+temp;|code-block|syntax|javascript|1177894|您可以使用以下命令将其设置为两行代码：|1177895|temp+=+x;+y+=+y+%2B+temp+-(x=y);|1177896|然后使用以下命令将其设置为一行：|1177897|x+=+x+%2B+y+-(y=x);|1177898|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]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Since the original solution is:

<pre><code>temp = x; y = x; x = temp;
</code></pre>

You can make it a two liner by using: 

<pre><code>temp = x; y = y + temp -(x=y);
</code></pre>

Then make it a one liner by using: 

<pre><code>x = x + y -(y=x);
</code></pre>

blocks|key|1393257|text|如果你稍微改变一下问题，询问两个汇编寄存器而不是变量，你也可以使用xchg操作作为一个选项，堆栈操作作为另一个选项。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1393258|entityMap^0|X|4|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

If you change a little the question to ask about 2 assembly registers instead of variables, you can use also the <code>xchg</code> operation as one option, and the stack operation as another one.

blocks|key|1393384|text|这里还有一个解决方案，但只有一个风险。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1393385|代码：|1393386|#include+<iostream>
#include+<conio.h>
void+main()
{

int+a+=10+,+b+=45;
*(&a%2B1+)+=+a;
a+=b;
b+=*(&a+%2B1);
}|code-block|syntax|javascript|1393387|位置a%2B1处的任何值都将被覆盖。|1393388|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|$]]

Here is one more solution but a single risk.

code:

<pre><code>#include &lt;iostream&gt;
#include &lt;conio.h&gt;
void main()
{

int a =10 , b =45;
*(&amp;a+1 ) = a;
a =b;
b =*(&amp;a +1);
}
</code></pre>

any value at location a+1 will be overridden.

blocks|key|1393678|text|当然，C%2B%2B的答案应该是std::swap。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1393679|但是，在下面的swap实现中也没有第三个变量|1393680|template+<typename+T>
void+swap+(T+&a,+T+&b)+{
++++std::pair<T+&,+T+&>(a,+b)+=+std::make_pair(b,+a);
}|code-block|syntax|javascript|1393681|或者，作为一行代码：|1393682|std::make_pair(std::ref(a),+std::ref(b))+=+std::make_pair(b,+a);|1393683|entityMap^0|C|9|0|7|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

Of course, the C++ answer should be <code>std::swap</code>.

However, there is also no third variable in the following implementation of <code>swap</code>:

<pre><code>template &lt;typename T&gt;
void swap (T &amp;a, T &amp;b) {
 std::pair&lt;T &amp;, T &amp;&gt;(a, b) = std::make_pair(b, a);
}
</code></pre>

Or, as a one-liner:

<pre><code>std::make_pair(std::ref(a), std::ref(b)) = std::make_pair(b, a);
</code></pre>

blocks|key|1393276|text|#include+<stdio.h>

int+main()
{
++++int+a,+b;
++++printf("Enter+A+:");
++++scanf("%25d",&a);
++++printf("Enter+B+:");
++++scanf("%25d",&b);
++++a+%5E=+b;
++++b+%5E=+a;
++++a+%5E=+b;
++++printf("\nValue+of+A=%25d+B=%25d+",a,b);
++++return+1;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1393277|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include &lt;stdio.h&gt;

int main()
{
 int a, b;
 printf("Enter A :");
 scanf("%d",&amp;a);
 printf("Enter B :");
 scanf("%d",&amp;b);
 a ^= b;
 b ^= a;
 a ^= b;
 printf("\nValue of A=%d B=%d ",a,b);
 return 1;
}
</code></pre>

blocks|key|1177985|text|这就是正确的XOR交换算法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177986|void+xorSwap+(int*+x,+int*+y)+{
+++if+(x+!=+y)+{+//ensure+that+memory+locations+are+different
++++++if+(*x+!=+*y)+{+//ensure+that+values+are+different
+++++++++*x+%5E=+*y;
+++++++++*y+%5E=+*x;
+++++++++*x+%5E=+*y;
++++++}
+++}
}|code-block|syntax|javascript|1177987|您必须确保内存位置不同，并且实际值也不同，因为A+XOR+A=0|1177988|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|$]]

that's the correct XOR swap algorithm

<pre><code>void xorSwap (int* x, int* y) {
 if (x != y) { //ensure that memory locations are different
 if (*x != *y) { //ensure that values are different
 *x ^= *y;
 *y ^= *x;
 *x ^= *y;
 }
 }
}
</code></pre>

you have to ensure that memory locations are different and also that the actual values are different because A XOR A = 0

blocks|key|1393552|text|您可以使用do....in+easy+way...within单行逻辑|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1393553|#include+<stdio.h>

int+main()
{
++++int+a,+b;
++++printf("Enter+A+:");
++++scanf("%25d",&a);
++++printf("Enter+B+:");
++++scanf("%25d",&b);
++++int+a+=+1,b+=+2;
++++a=a%5Eb%5E(b=a);
++++printf("\nValue+of+A=%25d+B=%25d+",a,b);

++++return+1;
}|code-block|syntax|javascript|1393554|或|1393555|#include+<stdio.h>

int+main()
{
++++int+a,+b;
++++printf("Enter+A+:");
++++scanf("%25d",&a);
++++printf("Enter+B+:");
++++scanf("%25d",&b);
++++int+a+=+1,b+=+2;
++++a=a%2Bb-(b=a);
++++printf("\nValue+of+A=%25d+B=%25d+",a,b);

++++return+1;
}|1393556|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You may do....in easy way...within one line Logic 

<pre><code>#include &lt;stdio.h&gt;

int main()
{
 int a, b;
 printf("Enter A :");
 scanf("%d",&amp;a);
 printf("Enter B :");
 scanf("%d",&amp;b);
 int a = 1,b = 2;
 a=a^b^(b=a);
 printf("\nValue of A=%d B=%d ",a,b);

 return 1;
}
</code></pre>

or

<pre><code>#include &lt;stdio.h&gt;

int main()
{
 int a, b;
 printf("Enter A :");
 scanf("%d",&amp;a);
 printf("Enter B :");
 scanf("%d",&amp;b);
 int a = 1,b = 2;
 a=a+b-(b=a);
 printf("\nValue of A=%d B=%d ",a,b);

 return 1;
}
</code></pre>

blocks|key|1178045|text|public+void+swapnumber(int+a,int+b){
++++a+=+a%2Bb-(b=a);
++++System.out.println("a+=+"%2Ba+%2B"+b=+"%2Bb);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1178046|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public void swapnumber(int a,int b){
 a = a+b-(b=a);
 System.out.println("a = "+a +" b= "+b);
}
</code></pre>

blocks|key|1393615|text|最好的答案是使用XOR，并且在一行中使用它将是很酷的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1393616|++++(x+%5E=+y),+(y+%5E=+x),+(x+%5E=+y);|code-block|syntax|javascript|1393617|x，y是变量，它们之间的逗号引入了序列点，因此它不依赖于编译器。干杯!|1393618|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|$]]

The best answer would be to use XOR and to use it in one line would be cool. 

<pre><code> (x ^= y), (y ^= x), (x ^= y);
</code></pre>

x,y are variables and the comma between them introduces the sequence points so it does not become compiler dependent. Cheers!

blocks|key|1178238|text|也许离题了，但如果你知道如何在两个不同的值之间交换单个变量，你也许能够做数组逻辑。每次运行这行代码时，它都会在1和2之间交换值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1178239|n+=+[2,+1][n+-+1]|code-block|syntax|javascript|1178240|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|$]]

Maybe off topic, but if you know what you are swapping a single variable between two different values, you may be able to do array logic. Each time this line of code is run, it will swap the value between 1 and 2.

<pre><code>n = [2, 1][n - 1]
</code></pre>

blocks|key|1393829|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1393830|std::tie(x,+y)+=+std::make_pair(y,+x);|code-block|syntax|javascript|1393831|或者在交换两个以上的变量时使用make_tuple：|1393832|std::tie(x,+y,+z)+=+std::make_tuple(y,+z,+x);|1393833|但我不确定内部std::tie是否使用临时变量！|1393834|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You could do:

<pre><code>std::tie(x, y) = std::make_pair(y, x);
</code></pre>

Or use make_tuple when swapping more than two variables:

<pre><code>std::tie(x, y, z) = std::make_tuple(y, z, x);
</code></pre>

But I'm not sure if internally std::tie uses a temporary variable or not!

blocks|key|1178341|text|在javascript中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1178342|function+swapInPlace(obj)+{
++++obj.x+%5E=+obj.y
++++obj.y+%5E=+obj.x
++++obj.x+%5E=+obj.y
}

function+swap(obj)+{
++++let+temp+=+obj.x
++++obj.x+=+obj.y
++++obj.y+=+temp
}|code-block|syntax|javascript|1178343|请注意这两个选项的执行时间。|1178344|通过运行这段代码，我测量了它。|1178345|console.time('swapInPlace')
swapInPlace({x:1,+y:2})
console.timeEnd('swapInPlace')+//+swapInPlace:+0.056884765625ms

console.time('swap')
swap({x:3,+y:6})
console.timeEnd('swap')++++++++//+swap:+0.01416015625ms|1178346|正如您所看到的(正如许多人所说的那样)，与使用temp变量的其他选项相比，原地交换(xor)需要更多的时间。|1178347|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

In javascript:

<pre><code>function swapInPlace(obj) {
 obj.x ^= obj.y
 obj.y ^= obj.x
 obj.x ^= obj.y
}

function swap(obj) {
 let temp = obj.x
 obj.x = obj.y
 obj.y = temp
}
</code></pre>

Be aware to execution time of both options. 

By run this code i measured it.

<pre><code>console.time('swapInPlace')
swapInPlace({x:1, y:2})
console.timeEnd('swapInPlace') // swapInPlace: 0.056884765625ms

console.time('swap')
swap({x:3, y:6})
console.timeEnd('swap') // swap: 0.01416015625ms
</code></pre>

As you can see (and as many said), swap in place (xor) take alot time than the other option using temp variable.

blocks|key|1178412|text|如果我们用"GO“lang进行并行处理..|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1178413|变量x=+100；变量y=+200；|1178414|swap1+:=+func(var+i)+{x=i}+swap2+:=+func(var+y)+{y=i}+Parallelize(swap1，swap2)|1178415|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

How about,If we do with parallel processing with &quot;GO&quot; lang..
var x = 100;
var y = 200;
swap1 := func(var i) { x = i }
swap2 := func(var y) { y = i }
Parallelize(swap1, swap2)

blocks|key|1177813|text|a+=+a+%2B+b+-+(b=a);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1177814|这很简单，但可能会引起警告。|unstyled|1177815|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>a = a + b - (b=a);
</code></pre>

It's very simple, but it may raise a warning.

blocks|key|1177923|text|在c语言中交换两个值的单行解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177924|a=(b=(a=a%2Bb,a-b),a-b);|code-block|syntax|javascript|1177925|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|$]]

single line solution for swapping two values in c language.

<pre><code>a=(b=(a=a+b,a-b),a-b);
</code></pre>

blocks|key|1393525|text|second_value+-=+first_value;
first_value+%2B=++second_value;
second_value+-=+first_value;
second_value+*=+-1;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1393526|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>second_value -= first_value;
first_value += second_value;
second_value -= first_value;
second_value *= -1;
</code></pre>

One of the very tricky questions asked in an interview.

Swap the values of two variables like <code>a=10</code> and <code>b=15</code>.

Generally to swap two variables values, we need 3rd variable like:

<pre><code>temp=a;
a=b;
b=temp;
</code></pre>

Now the requirement is, swap values of two variables without using 3rd variable.

Swapping two variable value without using third variable

面试中被问到的一个非常棘手的问题。交换两个变量的值，如a=10和b=15。通常，要交换两个变量值，我们需要第三个变量，如下所示：temp=a;a=b;b=temp;现在的要求是，交换两个变量的值，而不使用第三个变量。

问交换两个变量值而不使用第三个变量
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回答 22

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