blocks|key|1028874|text|开始吧：(a+%2B+b)+/+2+%2B+%7Ca+-+b%7C+/+2|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1028875|entityMap^0|4|P|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|$]]

Here we go: <code>(a + b) / 2 + |a - b| / 2</code>

blocks|key|814142|text|使用按位黑客|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814143|r+=+x+%5E+((x+%5E+y)+&+-(x+<+y));+//+max(x,+y)|code-block|syntax|javascript|814144|如果您知道INT_MIN+<=+x+-+y+<=+INT_MAX,，那么您可以使用以下代码，因为(x+-+y)只需要计算一次，因此速度更快。|offset|length|style|CODE|814145|r+=+x+-+((x+-+y)+&+((x+-+y)+>>+(sizeof(int)+*+CHAR_BIT+-+1)));+//+max(x,+y)|814146|来源：Bit+Twiddling+Hacks+by+Sean+Eron+Anderson|814147|entityMap|0|LINK|mutability|MUTABLE|url|http://www.graphics.stanford.edu/~seander/bithacks.html^0|0|0|5|S|1C|7|0|0|3|15|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$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]]|$1|O|3|P|5|6|7|16|8|@]|9|@$I|17|J|18|1|19]]|A|$]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

Use bitwise hacks

<pre><code>r = x ^ ((x ^ y) &amp; -(x &lt; y)); // max(x, y)
</code></pre>

If you know that <code>INT_MIN &lt;= x - y &lt;= INT_MAX,</code> then you can use the following, which is faster because <code>(x - y)</code> only needs to be evaluated once.

<pre><code>r = x - ((x - y) &amp; ((x - y) &gt;&gt; (sizeof(int) * CHAR_BIT - 1))); // max(x, y)
</code></pre>

Source : <a href="http://www.graphics.stanford.edu/~seander/bithacks.html" rel="noreferrer">Bit Twiddling Hacks by Sean Eron Anderson</a>

blocks|key|1029035|text|(sqrt(+a*a+%2B+b*b+-+2*a*b+)+%2B+a+%2B+b)+/+2|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1029036|这是基于与mike.dld's+solution相同的技术，但我在这里所做的事情并不那么“明显”。"abs“操作看起来像是在比较某些东西的符号，但我在这里利用的是sqrt()总是返回正平方根的事实，所以我将其平方(a-b)，将其完整地写出来，然后再次平方根，添加a%2Bb并除以2。|unstyled|offset|length|1029037|你会看到它总是有效的:例如，用户的例子10和5，你得到sqrt(100+%2B+25+-+100)+=5，然后将10和5相加得到20，除以2得到10。|1029038|如果我们使用9和11作为我们的数字，我们将得到(sqrt(121+%2B+81+-+198)+%2B+11+%2B+9)/2+=+(sqrt(4)+%2B+20)+/2=+22/2+=+11|1029039|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/4772786/3789665^0|0|5|J|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@$G|W|H|X|1|Y]]|A|$]]|$1|I|3|J|5|F|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<pre><code>(sqrt( a*a + b*b - 2*a*b ) + a + b) / 2
</code></pre>

This is based on the same technique as <a href="https://stackoverflow.com/a/4772786/3789665">mike.dld's solution</a>, but it is less "obvious" here what I am doing. An "abs" operation looks like you are comparing the sign of something but I here am taking advantage of the fact that sqrt() will always return you the positive square root so I am squaring (a-b) writing it out in full then square-rooting it again, adding a+b and dividing by 2.

You will see it always works: eg the user's example of 10 and 5 you get sqrt(100 + 25 - 100) = 5 then add 10 and 5 gives you 20 and divide by 2 gives you 10.

If we use 9 and 11 as our numbers we would get (sqrt(121 + 81 - 198) + 11 + 9)/2 = (sqrt(4) + 20) / 2 = 22/2 = 11

blocks|key|814672|text|最简单的答案如下。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814673|#include+<math.h>

int+Max(int+x,+int+y)
{
++++return+(float)(x+%2B+y)+/+2.0+%2B+abs((float)(x+-+y)+/+2);
}

int+Min(int+x,+int+y)
{
++++return+(float)(x+%2B+y)+/+2.0+-+abs((float)(x+-+y)+/+2);
}|code-block|syntax|javascript|814674|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|$]]

The simplest answer is below. 

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

int Max(int x, int y)
{
 return (float)(x + y) / 2.0 + abs((float)(x - y) / 2);
}

int Min(int x, int y)
{
 return (float)(x + y) / 2.0 - abs((float)(x - y) / 2);
}
</code></pre>

blocks|key|1029213|text|int+max(int+i,+int+j)+{
++++int+m+=+((i-j)+>>+31);
++++return+(m+&+j)+%2B+((~m)+&+i);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1029214|此解决方案避免了乘法。M将为0x00000000或0xffffffff|unstyled|1029215|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>int max(int i, int j) {
 int m = ((i-j) &gt;&gt; 31);
 return (m &amp; j) + ((~m) &amp; i);
}
</code></pre>

This solution avoids multiplication.
m will either be 0x00000000 or 0xffffffff

blocks|key|814396|text|以下是我认为我会如何做这项工作。它的可读性并不像你想要的那样好，但是当你从“我如何做X而不使用显而易见的方式做X”开始时，你必须有所期待。从理论上讲，这也放弃了一些可移植性，但你必须找到一个非常不寻常的系统才能看到问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814397|#define+BITS+(CHAR_BIT+*+sizeof(int)+-+1)

int+findmax(int+a,+int+b)+{+
++++int+rets[]+=+{a,+b};
++++return+rets[unsigned(a-b)>>BITS];
}|code-block|syntax|javascript|814398|这确实比问题中显示的有一些优势。首先，它计算正确的移位大小，而不是为32位整数硬编码。其次，对于大多数编译器，我们可以期望所有的乘法都在编译时发生，所以在运行时剩下的就是微不足道的位操作(减法和移位)，然后是加载和返回。简而言之，这几乎肯定是相当快的，即使在最小的微控制器上，原始的使用了必须在运行时发生的乘法，所以虽然在台式机上可能相当快，但在小型微控制器上通常会慢一些。|814399|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|$]]

Here's how I think I'd do the job. It's not as readable as you might like, but when you start with "how do I do X without using the obvious way of doing X, you have to kind of expect that.
In theory, this gives up some portability too, but you'd have to find a pretty unusual system to see a problem.

<pre><code>#define BITS (CHAR_BIT * sizeof(int) - 1)

int findmax(int a, int b) { 
 int rets[] = {a, b};
 return rets[unsigned(a-b)&gt;&gt;BITS];
}
</code></pre>

This does have some advantages over the one shown in the question. First of all, it calculates the correct size of shift, instead of being hard-coded for 32-bit ints. Second, with most compilers we can expect all the multiplication to happen at compile time, so all that's left at run time is trivial bit manipulation (subtract and shift) followed by a load and return. In short, this is almost certain to be pretty fast, even on the smallest microcontroller, where the original used multiplication that had to happen at run-time, so while it's probably pretty fast on a desktop machine, it'll often be quite a bit slower on a small microcontroller.

blocks|key|1028966|text|下面是这些行所做的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1028967|C是a-b。如果c为负，则为a|1028968|K是c的第32位，c是c的符号位(假设32位整数。如果在包含64位整数的平台上执行，此代码将无法工作)。它向右移动了31位，删除了最右边的31位，将符号位留在了最右边，然后用1对它进行and运算，以删除左侧的所有位(如果c为负数，将用1填充)。所以如果c是负的，k就是1，如果c是正的，k就是0。|1028969|
|1028970|如果c是0，这意味着a>=b，所以max是a+-0*c=a，如果c是1，这意味着a|1028971|总的来说，它只是使用差值的符号位，以避免使用大于或小于运算。老实说，说这段代码没有使用比较是有点愚蠢的。C是比较a和b的结果，代码只是没有使用比较运算符。您可以在许多汇编代码中执行类似的操作，只需减去这些数字，然后根据状态寄存器中设置的值进行跳转。|1028972|1028973|我还应该补充说，所有这些解决方案都假设这两个数字是整数。如果它们是浮点数、双精度数或更复杂的东西(BigInts、有理数等)。那么你真的需要使用一个比较运算符。比特技巧通常不会对这些人起作用。|1028974|1028975|1028976|1028977|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@]|9|@]|A|$]]|$1|L|3|G|5|6|7|Z|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|10|8|@]|9|@]|A|$]]|$1|O|3|G|5|6|7|11|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|S|$]]

Here's what those lines are doing:

c is a-b. if c is negative, a&lt;b.

k is 32nd bit of c which is the sign bit of c (assuming 32 bit integers. If done on a platform with 64 bit integers, this code will not work). It's shifted 31 bits to the right to remove the rightmost 31 bits leaving the sign bit in the right most place and then anding it with 1 to remove all the bits to the left (which will be filled with 1s if c is negative). So k will be 1 if c is negative and 0 if c is positive.

Then max = a - k * c. If c is 0, this means a>=b, so max is a - 0 * c = a. If c is 1, this means that a&lt;b and then a - 1 * c = a - (a - b) = a - a + b = b.

In the overall, it's just using the sign bit of the difference to avoid using greater than or less than operations. It's honestly a little silly to say that this code doesn't use a comparison. c is the result of comparing a and b. The code just doesn't use a comparison operator. You could do a similar thing in many assembly codes by just subtracting the numbers and then jumping based on the values set in the status register.

I should also add that all of these solutions are assuming that the two numbers are integers. If they are floats, doubles, or something more complicated (BigInts, Rational numbers, etc.) then you really have to use a comparison operator. Bit-tricks will not generally do for those.

blocks|key|814523|text|静态int+mymax(int+a，int+b)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814524|++++{
++++++++int[]+arr;
++++++++arr+=+new+int[3];
++++++++arr[0]+=+b;
++++++++arr[1]+=+a;
++++++++arr[2]+=+a;
++++++++return+arr[Math.Sign(a+-+b)+%2B+1];

++++}|code-block|syntax|javascript|814525|如果b>a，则(+a-b+)将为负，sign将返回-1，通过添加1，我们得到索引0，即b，如果为b=a，则a-b将为0，%2B1将给出1索引，因此无论我们返回a还是b，当a>b时，a-b将为正，sign将返回1，再加上1，我们将得到存储a的索引2。|814526|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|$]]

static int mymax(int a, int b)

<pre><code> {
 int[] arr;
 arr = new int[3];
 arr[0] = b;
 arr[1] = a;
 arr[2] = a;
 return arr[Math.Sign(a - b) + 1];

 }
</code></pre>

If b > a then (a-b) will be negative, sign will return -1, by adding 1 we get index 0 which is b, if b=a then a-b will be 0, +1 will give 1 index so it does not matter if we are returning a or b, when a > b then a-b will be positive and sign will return 1, adding 1 we get index 2 where a is stored.

blocks|key|814783|text|#include<stdio.h>
main()
{
++++++++int+num1,num2,diff;
++++++++printf("Enter+number+1+:+");
++++++++scanf("%25d",&num1);
++++++++printf("Enter+number+2+:+");
++++++++scanf("%25d",&num2);
++++++++diff=num1-num2;
++++++++num1=abs(diff);
++++++++num2=num1%2Bdiff;
++++++++if(num1==num2)
++++++++++++++++printf("Both+number+are+equal\n");
++++++++else+if(num2==0)
++++++++++++++++printf("Num2+>+Num1\n");
++++++++else
++++++++++++++++printf("Num1+>+Num2\n");
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|814784|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;
main()
{
 int num1,num2,diff;
 printf("Enter number 1 : ");
 scanf("%d",&amp;num1);
 printf("Enter number 2 : ");
 scanf("%d",&amp;num2);
 diff=num1-num2;
 num1=abs(diff);
 num2=num1+diff;
 if(num1==num2)
 printf("Both number are equal\n");
 else if(num2==0)
 printf("Num2 &gt; Num1\n");
 else
 printf("Num1 &gt; Num2\n");
}
</code></pre>

blocks|key|814824|text|//在C#中，可以使用数学库执行min或max函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814825|使用系统；|814826|类NumberComparator+{|814827|static+void+Main()
{

++++Console.Write("+write+the+first+number+to+compare:+");
++++double+first_Number+=+double.Parse(Console.ReadLine());

++++Console.Write("+write+the+second+number+to+compare:+");
++++double+second_Number+=+double.Parse(Console.ReadLine());

++++double+compare_Numbers+=+Math.Max(first_Number,+second_Number);
++++Console.Write("{0}+is+greater",compare_Numbers);

}|code-block|syntax|javascript|814828|}|814829|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

//In C# you can use math library to perform min or max function

using System;

class NumberComparator
{

<pre><code>static void Main()
{

 Console.Write(" write the first number to compare: ");
 double first_Number = double.Parse(Console.ReadLine());

 Console.Write(" write the second number to compare: ");
 double second_Number = double.Parse(Console.ReadLine());

 double compare_Numbers = Math.Max(first_Number, second_Number);
 Console.Write("{0} is greater",compare_Numbers);

}
</code></pre>

}

blocks|key|814768|text|问题中描述的逻辑可以解释为，如果第一个数字小于0，则减去0，否则从第一个数字中减去差值，得到第二个数字。我发现了另一个数学解决方案，我认为它更容易理解这个概念。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814769|将a和b视为给定的数字|814770|c=%7Ca/b%7C%2B1;
d=(c-1)/b;
smallest+number=+a+-+d*(a-b);|code-block|syntax|javascript|814771|同样，我们的想法是找到k，k等于0或1，然后将其与两个numbers.And的差相乘，最后从第一个数字中减去这个数字，得到两个数字中较小的一个。附注:此解决方案在第二个数字为零的情况下将失败|814772|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|$]]

The logic described in a problem can be explained as if 1st number is smaller then 0 will be subtracted else difference will be subtracted from 1st number to get 2nd number.
I found one more mathematical solution which I think is bit simpler to understand this concept.

Considering a and b as given numbers

<pre><code>c=|a/b|+1;
d=(c-1)/b;
smallest number= a - d*(a-b);
</code></pre>

Again,The idea is to find k which is wither 0 or 1 and multiply it with difference of two numbers.And finally this number should be subtracted from 1st number to yield the smaller of the two numbers.
P.S. this solution will fail in case 2nd number is zero

blocks|key|1029092|text|int+a=151;
int+b=121;
int+k=Math.abs(a-b);
int+j=+a%2Bb;
double+k1=(double)(k);
double+j1=+(double)+(j);
double+c=Math.ceil(k1/2)+%2B+Math.floor(j1/2);
int+c1=+(int)+(c);
System.out.println("+Max+value+=+"+%2B+c1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1029093|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>int a=151;
int b=121;
int k=Math.abs(a-b);
int j= a+b;
double k1=(double)(k);
double j1= (double) (j);
double c=Math.ceil(k1/2) + Math.floor(j1/2);
int c1= (int) (c);
System.out.println(" Max value = " + c1);
</code></pre>

blocks|key|814691|text|只有一种方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814692|+public+static+int+Min(int+a,+int+b)
++{
+++int+dif+=+(int)(((uint)(a+-+b))+>>+31);
+++return+a+*+dif+%2B+b+*+(1+-+dif);
++}|code-block|syntax|javascript|814693|和一个|814694|return+(a>=b)?b:a;|814695|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|$]]

There is one way

<pre><code> public static int Min(int a, int b)
 {
 int dif = (int)(((uint)(a - b)) &gt;&gt; 31);
 return a * dif + b * (1 - dif);
 }
</code></pre>

and one 

<pre><code>return (a&gt;=b)?b:a;
</code></pre>

blocks|key|814320|text|我想我们可以将这些数字与它们的逐位比较相乘，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814321|int+max=(a>b)*a%2B(a<=b)*b;|code-block|syntax|javascript|814322|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|$]]

Guess we can just multiply the numbers with their bitwise comparisons eg:

<pre><code>int max=(a&gt;b)*a+(a&lt;=b)*b;
</code></pre>

Find the maximum of two numbers. You should not use if-else or any other comparison operator. I found this question on online bulletin board, so i thought i should ask in StackOverflow

EXAMPLE
Input: 5, 10
Output: 10

I found this solution, can someone help me understand these lines of code

<pre><code>int getMax(int a, int b) { 
 int c = a - b; 
 int k = (c &gt;&gt; 31) &amp; 0x1; 
 int max = a - k * c; 
 return max; 
}
</code></pre>

Explain this snippet which finds the maximum of two integers without using if-else or any other comparison operator?

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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找到两个数字中的最大值。不应使用if-else或任何其他比较运算符。我在网上公告牌上发现了这个问题，所以我想我应该在StackOverflow上问一下示例输入: 5，10输出: 10我找到了这个解决方案，有人能帮我理解这几行代码吗int getMax(int a, int b) {      int c = a - b;      int k = (c >> 31) & 0x1;      int

问解释这段代码，它不使用if-else或任何其他比较运算符来查找两个整数的最大值？
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回答 14

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问解释这段代码，它不使用if-else或任何其他比较运算符来查找两个整数的最大值？EN

回答 14

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问解释这段代码，它不使用if-else或任何其他比较运算符来查找两个整数的最大值？
EN