blocks|key|1099901|text|由于ieee754将向-inf舍入指定为所需的舍入模式之一，因此我想您的问题的答案是肯定的。但也许您可以解释一下，如果编写编译器，您是想知道如何实现过程，还是想知道如何使用特定的编译器来执行操作？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1099902|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|$]]

Since IEEE-754 specifies round towards -inf as one of the required rounding modes I imagine that the answer to your question is yes. But perhaps you can explain whether you want to know how one would implement the procedure if one were writing the compiler, or to know how to use a particular compiler to perform the operation ?

blocks|key|1099946|text|可以更有效地提出一些分支自由的东西来根据符号纠正结果，因为分支是昂贵的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1099947|有关如何访问该标志，请参阅Hacker's+Delight中Chapter+2的第20ff页。|offset|length|1099948|entityMap|0|LINK|mutability|MUTABLE|url|http://www.hackersdelight.org/|1|http://www.hackersdelight.org/basics2.pdf^0|0|D|G|0|U|9|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@$D|R|E|S|1|T]|$D|U|E|V|1|W]]|A|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]|N|$5|I|J|K|A|$L|O]]]]

It could be more efficient to come up with something branch free to correct the result based on the sign, as branches are expensive.

See page 20ff of <a href="http://www.hackersdelight.org/basics2.pdf" rel="nofollow">Chapter 2</a> in <a href="http://www.hackersdelight.org/" rel="nofollow">Hacker's Delight</a> on how to access the sign.

blocks|key|1100029|text|我已经编写了一个测试程序来对这里提出的想法进行基准测试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1100030|#include+<stdio.h>
#include+<stdlib.h>
#include+<math.h>
#include+<windows.h>

#define+N+10000000
#define+M+100

int+dividends[N],+divisors[N],+results[N];

__forceinline+int+floordiv_signcheck(int+a,+int+b)
{
++++return+(a<0+?+a-(b-1)+:+a)+/+b;
}

__forceinline+int+floordiv_signcheck2(int+a,+int+b)
{
++++return+(a+-+(a<0+?+b-1+:+0))+/+b;
}

__forceinline+int+floordiv_signmultiply(int+a,+int+b)
{
++++return+(a+%2B+(a>>(sizeof(a)*8-1))*(b-1))+/+b;
}

__forceinline+int+floordiv_floatingpoint(int+a,+int+b)
{
++++//+I+imagine+that+the+call+to+floor+can+be+replaced+to+a+cast
++++//+if+you+can+get+FPU+rounding+control+to+work+(I+couldn't).
++++return+floor((double)a+/+b);
}

void+main()
{
++++for+(int+i=0;+i<N;+i%2B%2B)
++++{
++++++++dividends[i]+=+rand();
++++++++do
++++++++++++divisors[i]+=+rand();
++++++++while+(divisors[i]==0);
++++}

++++LARGE_INTEGER+t0,+t1;

++++QueryPerformanceCounter(&t0);
++++for+(int+j=0;+j<M;+j%2B%2B)
++++++++for+(int+i=0;+i<N;+i%2B%2B)
++++++++++++results[i]+=+floordiv_signcheck(dividends[i],+divisors[i]);
++++QueryPerformanceCounter(&t1);
++++printf("signcheck++++:+%259llu\n",+t1.QuadPart-t0.QuadPart);

++++QueryPerformanceCounter(&t0);
++++for+(int+j=0;+j<M;+j%2B%2B)
++++++++for+(int+i=0;+i<N;+i%2B%2B)
++++++++++++results[i]+=+floordiv_signcheck2(dividends[i],+divisors[i]);
++++QueryPerformanceCounter(&t1);
++++printf("signcheck2+++:+%259llu\n",+t1.QuadPart-t0.QuadPart);

++++QueryPerformanceCounter(&t0);
++++for+(int+j=0;+j<M;+j%2B%2B)
++++++++for+(int+i=0;+i<N;+i%2B%2B)
++++++++++++results[i]+=+floordiv_signmultiply(dividends[i],+divisors[i]);
++++QueryPerformanceCounter(&t1);
++++printf("signmultiply+:+%259llu\n",+t1.QuadPart-t0.QuadPart);

++++QueryPerformanceCounter(&t0);
++++for+(int+j=0;+j<M;+j%2B%2B)
++++++++for+(int+i=0;+i<N;+i%2B%2B)
++++++++++++results[i]+=+floordiv_floatingpoint(dividends[i],+divisors[i]);
++++QueryPerformanceCounter(&t1);
++++printf("floatingpoint:+%259llu\n",+t1.QuadPart-t0.QuadPart);
}|code-block|syntax|javascript|1100031|结果：|1100032|signcheck++++:++61458768
signcheck2+++:++61284370
signmultiply+:++61625076
floatingpoint:+287315364|1100033|因此，根据我的结果，检查标志是最快的：|1100034|(a+-+(a<0+?+b-1+:+0))+/+b|1100035|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|$]]

I've written a test program to benchmark the ideas presented here:

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

#define N 10000000
#define M 100

int dividends[N], divisors[N], results[N];

__forceinline int floordiv_signcheck(int a, int b)
{
 return (a&lt;0 ? a-(b-1) : a) / b;
}

__forceinline int floordiv_signcheck2(int a, int b)
{
 return (a - (a&lt;0 ? b-1 : 0)) / b;
}

__forceinline int floordiv_signmultiply(int a, int b)
{
 return (a + (a&gt;&gt;(sizeof(a)*8-1))*(b-1)) / b;
}

__forceinline int floordiv_floatingpoint(int a, int b)
{
 // I imagine that the call to floor can be replaced to a cast
 // if you can get FPU rounding control to work (I couldn't).
 return floor((double)a / b);
}

void main()
{
 for (int i=0; i&lt;N; i++)
 {
 dividends[i] = rand();
 do
 divisors[i] = rand();
 while (divisors[i]==0);
 }

 LARGE_INTEGER t0, t1;

 QueryPerformanceCounter(&amp;t0);
 for (int j=0; j&lt;M; j++)
 for (int i=0; i&lt;N; i++)
 results[i] = floordiv_signcheck(dividends[i], divisors[i]);
 QueryPerformanceCounter(&amp;t1);
 printf("signcheck : %9llu\n", t1.QuadPart-t0.QuadPart);

 QueryPerformanceCounter(&amp;t0);
 for (int j=0; j&lt;M; j++)
 for (int i=0; i&lt;N; i++)
 results[i] = floordiv_signcheck2(dividends[i], divisors[i]);
 QueryPerformanceCounter(&amp;t1);
 printf("signcheck2 : %9llu\n", t1.QuadPart-t0.QuadPart);

 QueryPerformanceCounter(&amp;t0);
 for (int j=0; j&lt;M; j++)
 for (int i=0; i&lt;N; i++)
 results[i] = floordiv_signmultiply(dividends[i], divisors[i]);
 QueryPerformanceCounter(&amp;t1);
 printf("signmultiply : %9llu\n", t1.QuadPart-t0.QuadPart);

 QueryPerformanceCounter(&amp;t0);
 for (int j=0; j&lt;M; j++)
 for (int i=0; i&lt;N; i++)
 results[i] = floordiv_floatingpoint(dividends[i], divisors[i]);
 QueryPerformanceCounter(&amp;t1);
 printf("floatingpoint: %9llu\n", t1.QuadPart-t0.QuadPart);
}
</code></pre>

Results: 

<pre><code>signcheck : 61458768
signcheck2 : 61284370
signmultiply : 61625076
floatingpoint: 287315364
</code></pre>

So, according to my results, checking the sign is the fastest:

<pre><code>(a - (a&lt;0 ? b-1 : 0)) / b
</code></pre>

blocks|key|1100109|text|只需注意一点:当涉及到2的幂时，x86+sar指令会执行floored除法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1100110|entityMap^0|K|3|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|$]]

Just a note: the x86 <code>sar</code> instruction performs floored division when it comes to powers of two.

Floored division is when the result is always floored down (towards −∞), not towards 0:

<img src="https://i.stack.imgur.com/ME0ZR.png" alt="division types">

Is it possible to efficiently implement floored or euclidean integer division in C/C++?

(the obvious solution is to check the dividend's sign)

Efficiently implementing floored / euclidean integer division

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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Floored除法是指结果总是朝下(朝向−∞)，而不是朝向0：​​是否有可能在C/C++中高效地实现floored或欧几里得整数除法？(显而易见的解决方案是检查红利的符号)

问高效实现floored / euclidean整数除法
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