最快浮点数取绝对值
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做视频算法10多年,经常要算绝对值,整数的绝对值有快速算法,但浮点数的绝对值没看到有快速算法,经常不段发现,得到如下浮点数的快速算法:
快6倍多,
#include <Windows.h>#include <iostream>#include <stdio.h>#include <stdlib.h>#include <math.h>union funion{float fnum;BYTE bnum[4];} ;float fabs_11(float x){return (x < 0) ? -x : x;}void main(void){int iTime;float ret;union funion tmp1,tmp2;float x =- 2110000.0001;iTime = GetTickCount();for(; x < 2210000.987;x+=0.3){tmp1.fnum = x;//printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp1.fnum,tmp1.bnum[3],tmp1.bnum[2],tmp1.bnum[1],tmp1.bnum[0]);// printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp2.fnum,tmp2.bnum[3],tmp2.bnum[2],tmp2.bnum[1],tmp2.bnum[0]);tmp1.bnum[3] = tmp1.bnum[3]&0x7f;//printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp1.fnum,tmp1.bnum[3],tmp1.bnum[2],tmp1.bnum[1],tmp1.bnum[0]);// printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp2.fnum,tmp2.bnum[3],tmp2.bnum[2],tmp2.bnum[1],tmp2.bnum[0]);}printf("gai time = %d\n",GetTickCount() - iTime);iTime = GetTickCount();x =- 2110000.0001;for(; x < 2210000.987;x+=0.3){ fabs_11(x);}printf("old time = %d\n",GetTickCount() - iTime);}
如果浮点数是double,将如何生成呢?
请大家先想想,后看答案。
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#include <Windows.h>#include <iostream>#include <stdio.h>#include <stdlib.h>#include <math.h>union funion{double fnum;BYTE bnum[8];} ;double fabs_11(double x){return (x < 0) ? -x : x;}void main(void){int iTime;double ret;union funion tmp1,tmp2;double x =- 2110000.0001;iTime = GetTickCount();for(; x < 2210000.987;x+=0.3){tmp1.fnum = x;printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp1.fnum,tmp1.bnum[3],tmp1.bnum[2],tmp1.bnum[1],tmp1.bnum[0]);// printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp2.fnum,tmp2.bnum[3],tmp2.bnum[2],tmp2.bnum[1],tmp2.bnum[0]);tmp1.bnum[7] = tmp1.bnum[7]&0x7f;printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp1.fnum,tmp1.bnum[3],tmp1.bnum[2],tmp1.bnum[1],tmp1.bnum[0]);// printf("%f\t%2.2x%2.2x%2.2x%2.2x\n",tmp2.fnum,tmp2.bnum[3],tmp2.bnum[2],tmp2.bnum[1],tmp2.bnum[0]);}printf("gai time = %d\n",GetTickCount() - iTime);iTime = GetTickCount();x =- 2110000.0001;for(; x < 2210000.987;x+=0.3){ fabs_11(x);}printf("old time = %d\n",GetTickCount() - iTime);}
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