blocks|key|1245928|text|有一些技巧可以找到两个数字中的最小值或最大值，因此您可以使用这些技巧来查找min(max(x,+0),+some_maximum)。来自here|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1245929|y+%5E+((x+%5E+y)+&+-(x+<+y));+//+min(x,+y)
x+%5E+((x+%5E+y)+&+-(x+<+y));+//+max(x,+y)|code-block|syntax|javascript|1245930|但是，正如源代码所述，使用常规方法可能会更快，尽管分支|1245931|entityMap|0|LINK|mutability|MUTABLE|url|http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax^0|11|S|1W|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

There are bit-tricks to find the minimum or maximum of two numbers, so you could use those to find <code>min(max(x, 0), some_maximum)</code>. From <a href="http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax" rel="noreferrer">here</a>:

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

As the source states though, it's probably faster to do it the normal way, despite the branch

blocks|key|1245848|text|使用三元运算符:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1245849|return+x+<+0+?+0+:+x+>+some_maximum+?+:+some_maximum+:+x;|code-block|syntax|javascript|1245850|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|$]]

Using the ternary operator :)

<pre><code>return x &lt; 0 ? 0 : x &gt; some_maximum ? : some_maximum : x;
</code></pre>

blocks|key|1030821|text|取决于您的体系结构。至少对于ARM，编译器可能会发出有条件地执行的指令，并且生成的机器码不会包含分支。不过，我想不出在C程序中明确这一点的好方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1030822|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Depends on your architecture. For ARM, at least, the compiler would probably emit conditionally executed instructions and the resulting machine code wouldn't contain a branch. I can't think of a good way to make that explicit in the C program though.

blocks|key|1030967|text|如果可以限制为2的幂(不包括在内)，那么只需使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1030968|int+newx+=+x+&+((highest+power+of+2)+-+1)|offset|length|style|CODE|1030969|不太确定是处理(如果x<0情况)还是一般情况(x+|1030970|1030971|entityMap^0|0|0|15|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@$D|O|E|P|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Q|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|L|$]]

If it's possible to limit to powers of 2 (non inclusive), then just go with 

<code>int newx = x &amp; ((highest power of 2) - 1)</code>

not quite sure to handle the (if x &lt; 0 case) or the generic (x &lt; n case)

blocks|key|1245987|text|对于未来类似的问题，bit+hack页面可能会有用：http://graphics.stanford.edu/~seander/bithacks.html。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1245988|由于min和max的bithack已经发布，这里是一个不同的版本：|1245989|//+CHAR_BIT+is+number+of+bits+per+byte.
//+sign+=+1+if+x+<+0,+sign+=+0+otherwise+(according+to+the+page+above)
int+sign+=+(int)((unsigned+int)((int)x)+>>+(sizeof(int)+*+CHAR_BIT+-+1));

int+y+=+(1-sign)*x;+//+if+x+<+0,+then+y+=+0,+else+y+=+x.

//+Depending+on+arch,+the+below+_might_+cause+a+branch.
//+(on+x64+it+does+not+cause+a+branch,+not+sure+about+MIPS)

int+z+=+!(y/some_maximum);+//+if+0+<=+y+<+some_maximum,+z+=+1,+else+z+=+0

int+ret+=+z*y+%2B+(1-z)*some_maximum;+//+if+z+=1,+then+ret+=+y;+else+ret+=+some_maximum.

return+ret;|code-block|syntax|javascript|1245990|我只是尝试了一下，它在我的几个测试用例中都有效。|1245991|这是我的计算机(英特尔拱门)的汇编代码，它没有显示分支。|1245992|int+cap(int+x)
{
00F013A0++push++++++++ebp++
00F013A1++mov+++++++++ebp,esp+
00F013A3++sub+++++++++esp,0FCh+
00F013A9++push++++++++ebx++
00F013AA++push++++++++esi++
00F013AB++push++++++++edi++
00F013AC++lea+++++++++edi,[ebp-0FCh]+
00F013B2++mov+++++++++ecx,3Fh+
00F013B7++mov+++++++++eax,0CCCCCCCCh+
00F013BC++rep+stos++++dword+ptr+es:[edi]+

++++int+some_maximum+=+100;

00F013BE++mov+++++++++dword+ptr+[some_maximum],64h+

++++//+CHAR_BIT+is+number+of+bits+per+byte.+
++++//+sign+=+1+if+x+<+0,+sign+=+0+otherwise+(according+to+the+page+above)+
++++int+sign+=+(int)((unsigned+int)((int)x)+>>+(sizeof(int)+*+CHAR_BIT+-+1));+

00F013C5++mov+++++++++eax,dword+ptr+[x]+
00F013C8++shr+++++++++eax,1Fh+
00F013CB++mov+++++++++dword+ptr+[sign],eax+

++++int+y+=+(1-sign)*x;+//+if+x+<+0,+then+y+=+0,+else+y+=+x.+

00F013CE++mov+++++++++eax,1+
00F013D3++sub+++++++++eax,dword+ptr+[sign]+
00F013D6++imul++++++++eax,dword+ptr+[x]+
00F013DA++mov+++++++++dword+ptr+[y],eax+

++++//+Depending+on+arch,+the+below+_might_+cause+a+branch.+
++++//+(on+x64+it+does+not+cause+a+branch,+not+sure+about+MIPS)+

++++int+z+=+!(y/some_maximum);+//+if+0+<=+y+<+some_maximum,+z+=+1,+else+z+=+0+

00F013DD++mov+++++++++eax,dword+ptr+[y]+
00F013E0++cdq++++++++++++++
00F013E1++idiv++++++++eax,dword+ptr+[some_maximum]+
00F013E4++neg+++++++++eax++
00F013E6++sbb+++++++++eax,eax+
00F013E8++add+++++++++eax,1+
00F013EB++mov+++++++++dword+ptr+[z],eax+

++++int+ret+=+z*y+%2B+(1-z)*some_maximum;+//+if+z+=1,+then+ret+=+y;+else+ret+=+some_maximum.+

00F013EE++mov+++++++++eax,dword+ptr+[z]+
00F013F1++imul++++++++eax,dword+ptr+[y]+
00F013F5++mov+++++++++ecx,1+
00F013FA++sub+++++++++ecx,dword+ptr+[z]+
00F013FD++imul++++++++ecx,dword+ptr+[some_maximum]+
00F01401++add+++++++++eax,ecx+
00F01403++mov+++++++++dword+ptr+[ret],eax+

++++return+ret;+

00F01406++mov+++++++++eax,dword+ptr+[ret]+
}
00F01409++pop+++++++++edi++
00F0140A++pop+++++++++esi++
00F0140B++pop+++++++++ebx++
00F0140C++mov+++++++++esp,ebp+
00F0140E++pop+++++++++ebp++
00F0140F++ret++++++++++++++|1245993|entityMap|0|LINK|mutability|MUTABLE|url|http://graphics.stanford.edu/~seander/bithacks.html^0|Q|1F|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|C|$]]|$1|O|3|P|5|H|7|16|8|@]|9|@]|C|$I|J]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

For future problems like this, the bit hack page might be useful: <a href="http://graphics.stanford.edu/~seander/bithacks.html" rel="nofollow noreferrer">http://graphics.stanford.edu/~seander/bithacks.html</a>. 

Since the bithack for min and max was already posted, here is a different one:

<pre><code>// CHAR_BIT is number of bits per byte.
// sign = 1 if x &lt; 0, sign = 0 otherwise (according to the page above)
int sign = (int)((unsigned int)((int)x) &gt;&gt; (sizeof(int) * CHAR_BIT - 1));

int y = (1-sign)*x; // if x &lt; 0, then y = 0, else y = x.

// Depending on arch, the below _might_ cause a branch.
// (on x64 it does not cause a branch, not sure about MIPS)

int z = !(y/some_maximum); // if 0 &lt;= y &lt; some_maximum, z = 1, else z = 0

int ret = z*y + (1-z)*some_maximum; // if z =1, then ret = y; else ret = some_maximum.

return ret;
</code></pre>

I just tried it out, and it worked for the few test cases i had.

Here is the assembly code from my computer (intel arch) which shows no branches.

<pre><code>int cap(int x)
{
00F013A0 push ebp 
00F013A1 mov ebp,esp 
00F013A3 sub esp,0FCh 
00F013A9 push ebx 
00F013AA push esi 
00F013AB push edi 
00F013AC lea edi,[ebp-0FCh] 
00F013B2 mov ecx,3Fh 
00F013B7 mov eax,0CCCCCCCCh 
00F013BC rep stos dword ptr es:[edi] 

 int some_maximum = 100;

00F013BE mov dword ptr [some_maximum],64h 

 // CHAR_BIT is number of bits per byte. 
 // sign = 1 if x &lt; 0, sign = 0 otherwise (according to the page above) 
 int sign = (int)((unsigned int)((int)x) &gt;&gt; (sizeof(int) * CHAR_BIT - 1)); 

00F013C5 mov eax,dword ptr [x] 
00F013C8 shr eax,1Fh 
00F013CB mov dword ptr [sign],eax 

 int y = (1-sign)*x; // if x &lt; 0, then y = 0, else y = x. 

00F013CE mov eax,1 
00F013D3 sub eax,dword ptr [sign] 
00F013D6 imul eax,dword ptr [x] 
00F013DA mov dword ptr [y],eax 

 // Depending on arch, the below _might_ cause a branch. 
 // (on x64 it does not cause a branch, not sure about MIPS) 

 int z = !(y/some_maximum); // if 0 &lt;= y &lt; some_maximum, z = 1, else z = 0 

00F013DD mov eax,dword ptr [y] 
00F013E0 cdq 
00F013E1 idiv eax,dword ptr [some_maximum] 
00F013E4 neg eax 
00F013E6 sbb eax,eax 
00F013E8 add eax,1 
00F013EB mov dword ptr [z],eax 

 int ret = z*y + (1-z)*some_maximum; // if z =1, then ret = y; else ret = some_maximum. 

00F013EE mov eax,dword ptr [z] 
00F013F1 imul eax,dword ptr [y] 
00F013F5 mov ecx,1 
00F013FA sub ecx,dword ptr [z] 
00F013FD imul ecx,dword ptr [some_maximum] 
00F01401 add eax,ecx 
00F01403 mov dword ptr [ret],eax 

 return ret; 

00F01406 mov eax,dword ptr [ret] 
}
00F01409 pop edi 
00F0140A pop esi 
00F0140B pop ebx 
00F0140C mov esp,ebp 
00F0140E pop ebp 
00F0140F ret 
</code></pre>

blocks|key|1030892|text|x+=+min(max(x,0),100);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1030893|分支很好地隐藏在具有普通名称的函数中。|unstyled|1030894|建议创建一个clip_by模板。|1030895|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>x = min(max(x,0),100);
</code></pre>

The branching is hidden away nicely inside functions with normal names.

Suggesting to create a clip_by template.

blocks|key|1245909|text|x+=+++((int)(x+>+some_maximum))+*+some_maximum+
++++%2B+((int)(x+>+0+&&+x+<=+some_maximum))+*+x;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1245910|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>x = ((int)(x &gt; some_maximum)) * some_maximum 
 + ((int)(x &gt; 0 &amp;&amp; x &lt;= some_maximum)) * x;
</code></pre>

Just out of curiosity. If I have something like:

<pre><code>if(x &lt; 0)
 x = 0;
if(x &gt; some_maximum)
 x = some_maximum;

return x;
</code></pre>

Is there a way to not branch? This is c++.

Addendum: I mean no branch instructions in the assembly. It's a MIPS architecture.

Is there a way to limit an integer value to a certain range without branching?

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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只是出于好奇。如果我有类似这样的东西：if(x < 0) x = 0;if(x > some_maximum) x = some_maximum;return x;有没有不分支的方法？这是c++。附录:我的意思是在程序集中没有分支指令。这是一个MIPS架构。

问有没有一种方法可以将整数值限制在某个范围内而不进行分支？
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回答 7

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问有没有一种方法可以将整数值限制在某个范围内而不进行分支？
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