浮点算法很慢 [英] float algorithm is slow
问题描述
浮动百分比;
for（j = 0; j <10000000; j ++）
{
percentage = sinf（频率* j * 2 * 3.14159 / sampleFreq）;
缓冲[totalBytes] = ceilf（音量*百分比）+音量;
totalBytes ++;
}
因为浮点变量，上面的循环在Linux机器上需要2秒c或c ++
。有没有人有办法减少时间？
谢谢，
文斐
Wenfei写道：
浮动百分比;
（j = 0; j <10000000; j ++ ）{
百分比= sinf（频率* j * 2 * 3.14159 / sampleFreq）;
缓冲区[totalBytes] = ceilf（音量*百分比）+音量;
totalBytes ++;
}
因为float变量，上面的循环在Linux机器上需要2秒c或c ++
。有没有人有减少时间的解决方案？
cat main.c
#include< stdlib.h>
#include< math.h>
int main （int argc，char * argv []）{
const
size_t n = 10000000;
float buffer [n] ;
size_t totalBytes = 0;
const
float_t frequency = 1.0;
const
float_t sampleFreq = 1.0;
const
float_t pi = 3.14159265358979323846;
const
float_t volume = 1.0;
for（size_t j = 0; j< n; ++ j）{
float_t percentage = sinf（频率* j * 2 * pi / sampleFreq）;
buffer [totalBytes] = ceilf（音量*百分比）+音量;
totalBytes ++;
}
返回0;
}
gcc -Wall -std = c99 -pedantic -O2 -o main main.c -lm
时间./main
3.694u 0.258s 0：03.92 100.5％0 + 0k 0 + 0io 0pf + 0w
文斐写道：浮动百分比;
（j = 0; j<千万; j ++）
{百分比= sinf（频率* j * 2 * 3.14159 / sampleFreq）;
缓冲区[totalBytes] = ceilf（音量*百分比）+音量;
totalBytes ++;
}
因为float变量，上面的循环在Linux机器上的c或c ++中需要2秒钟。有没有人有减少时间的解决方案？
是的：将迭代次数从10000000更改为0，并且
代码几乎可以肯定运行得更快。
换句话说，这种微基准测试并不是非常有用的。你真的想做什么？
-
Eric Sosman
es ***** @ acm-dot-org.inva lid
>浮动百分比;
for（j = 0; j <10000000; j ++）
{
percentage = sinf（frequency * j * 2 * 3.14159 / sampleFreq）;
buffer [totalBytes] = ceilf（音量*百分比）+音量;
totalBytes ++;
}
因为浮点变量，
我的猜测是，如果你获得浮动变量的RID，那么它将需要大约相同的时间：
for（j = 0; j <10000000; j ++ ）
{
buffer [totalBytes] = ceilf（volume * sinf（frequency * j * 2 * 3.14159 / sampleFreq））+ volume;
totalBytes ++;
}
在Linux机器上，上述循环在c或c ++中花费2秒钟。有没有人有减少时间的解决方案？
你还没有证明为什么花两秒钟是个问题。
Cut减少迭代次数？获得更快的机器？
以双倍计算可能会使它更快（尽管在
Intel * 86上它可能不会）。
Gordon L. Burditt
float percentage;
for (j = 0; j < 10000000; j++)
{
percentage = sinf(frequency * j * 2 * 3.14159 / sampleFreq );
buffer[totalBytes] =ceilf(volume * percentage) + volume;
totalBytes++;
}
Because the float variable, the above loop take 2 seconds in c or c++
on Linux machine. Does anybody has a solution to reduce the time?
Thanks,
Wenfei
Wenfei wrote:
float percentage;
for (j = 0; j < 10000000; j++) {
percentage = sinf(frequency * j * 2 * 3.14159 / sampleFreq );
buffer[totalBytes] =ceilf(volume * percentage) + volume;
totalBytes++;
}
Because the float variable, the above loop take 2 seconds in c or c++
on Linux machine. Does anybody has a solution to reduce the time? cat main.c #include <stdlib.h>
#include <math.h>
int main(int argc, char* argv[]) {
const
size_t n = 10000000;
float buffer[n];
size_t totalBytes = 0;
const
float_t frequency = 1.0;
const
float_t sampleFreq = 1.0;
const
float_t pi = 3.14159265358979323846;
const
float_t volume = 1.0;
for (size_t j = 0; j < n; ++j) {
float_t percentage = sinf(frequency*j*2*pi/sampleFreq);
buffer[totalBytes] =ceilf(volume*percentage) + volume;
totalBytes++;
}
return 0;
}
gcc -Wall -std=c99 -pedantic -O2 -o main main.c -lm
time ./main
3.694u 0.258s 0:03.92 100.5% 0+0k 0+0io 0pf+0w
Wenfei wrote:float percentage;
for (j = 0; j < 10000000; j++)
{
percentage = sinf(frequency * j * 2 * 3.14159 / sampleFreq );
buffer[totalBytes] =ceilf(volume * percentage) + volume;
totalBytes++;
}
Because the float variable, the above loop take 2 seconds in c or c++
on Linux machine. Does anybody has a solution to reduce the time?
Yes: Change the iteration count from 10000000 to 0, and
the code will almost certainly run faster.
In other words, micro-benchmarks of this sort are not
very informative. What are you really trying to do?
--
Eric Sosman
es*****@acm-dot-org.invalid
>float percentage;
for (j = 0; j < 10000000; j++)
{
percentage = sinf(frequency * j * 2 * 3.14159 / sampleFreq );
buffer[totalBytes] =ceilf(volume * percentage) + volume;
totalBytes++;
}
Because the float variable,
My guess is that if you GOT RID OF the float variable, it would
take about the same time:
for (j = 0; j < 10000000; j++) {
buffer[totalBytes] =ceilf(volume * sinf(frequency * j * 2 * 3.14159 / sampleFreq )) + volume;
totalBytes++;
}
the above loop take 2 seconds in c or c++
on Linux machine. Does anybody has a solution to reduce the time?
You haven''t demonstrated why taking two seconds is a problem yet.
Cut down the number of iterations? Get a faster machine?
Doing the calculation in double might make it faster (although on
Intel *86 it probably won''t).
Gordon L. Burditt
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