>>加速分裂 [英] >> to accelerate division
问题描述
我们通常使用<< n作为一种更便宜的乘法选项，当它的
因子为
2 ^ n对，
即使是非2 ^ n说
ix24 ==> ix（16 + 8）==> （i<< 4）+（i<<<<<<<<<<<<<<<<<<<<<<<<<<">>
类似地怀疑是否有更便宜的选择i / 24 ==> i /（16 + 8）==> ????????
或i / 72 .....
还有其他更快的算术选择，如>>和<<
for
div n乘法
fORDGE
we usually use <<n as a cheaper option to multiplication when its
factor of
2^n right,
even for non 2^n say
ix24 ==> ix(16+8) ==> (i<<4)+(i<<3)
similarly divison 2^n is >>n
doubt is there a cheaper option to i/24 ==> i/(16+8) ==> ????????
or i/72.....
also are there other faster arithmetic alternatives like the >> and <<
for
div n multiplication
fORDGE
推荐答案
好吧，你可以稍微减少它，但是分歧想要坚持
...
i /（24 ）== i /（6 * 4）== i / 6/4 == i /（3 * 2）/ 4 == i / 3/2/2/2
==（ i / 3）>> 1>> 1>> 1
i / 24 ==（i>> 3）/ 3
如果你能想出一种表达方式i / 3比特移位，
然后你有所改善。
但是我怀疑是否有办法，没有完全分裂。
-Jh
Well, you could reduce it somewhat, but the divide wants to stick
around...
i/(24) == i/(6*4) == i/6/4 == i/(3*2)/4 == i/3/2/2/2
== (i / 3) >> 1 >> 1 >> 1
i / 24 == (i >> 3) / 3
If you can figure out a way of expressing i / 3 as bit shifts,
then you have an improvement.
But I am doubting there is a way, without doing the full divide.
-Jh
< posted&邮寄>
请记住，您的编译器可能会将所有这些语句
转换为相同的代码：
i>> 1;
i / = 2;
i = i / 2;
.... so你不会指望任何性能差异。同样地，
跟随应该生成完全相同的代码：
i<< 1;
i * = 2;
i = i * 2;
此外，根据硬件，''i * 24''可能比''（i<< 4）+
（i<<< 3）''...假设您的机器没有将代码优化到
同样的事情......
fo****@gmail.com 写道：
<posted & mailed>
Keep in mind that your compiler will probably turn all of these statements
into the same code:
i>>1;
i/=2;
i=i/2;
.... so you wouldn''t expect any performance difference. Likewise, the
following should all generate exactly the same code:
i<<1;
i*=2;
i=i*2;
Also, depending on the hardware, ''i * 24'' may be faster than ''(i<<4) +
(i<<3)'' ... assuming that your machine didn''t optimize the code into the
same thing...
fo****@gmail.com wrote:
我们通常使用<< n作为一种更便宜的乘法选择，当它的因素为2 ^ n时，即使对于非2 ^ n说
/>
ix24 ==> ix（16 + 8）==> （i<<<<<<<<<<<<<<<<<<<<<<<<<<<">} / 24 ==> i /（16 + 8）==> ????????
或i / 72 .....
还有其他更快的算术选择，如>>和<<
div n乘法
fORDGE
we usually use <<n as a cheaper option to multiplication when its
factor of
2^n right,
even for non 2^n say
ix24 ==> ix(16+8) ==> (i<<4)+(i<<3)
similarly divison 2^n is >>n
doubt is there a cheaper option to i/24 ==> i/(16+8) ==> ????????
or i/72.....
also are there other faster arithmetic alternatives like the >> and <<
for
div n multiplication
fORDGE
-
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" Mysidia" <我的***** @ gmail.com>写道：
"Mysidia" <my*****@gmail.com> writes:
如果你能想出一种表达i / 3作为位移的方法，那么你就有了改进。
但我怀疑是否存在一种方式，没有完全分裂。
If you can figure out a way of expressing i / 3 as bit shifts,
then you have an improvement.
But I am doubting there is a way, without doing the full divide.
你可以实现i / 3作为乘法，然后是一点
shift。在某些情况下，这可能是一种改进。但是，如果它是
是一个改进，你的编译器可能已经为你做了。
见Warren，Jr。的第10章，_Hacker'' s Delight_，Addison-Wesley
2003，ISBN 0-201-91465-4，详情。
-
Ben Pfaff
电子邮件： bl*@cs.stanford.edu
web： http://benpfaff.org
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