绝望了，INTEL编程大赛第一题。

[yyc520]

为什么读取整个文件的开销要比读取整个文件且求出最大值的开销还要大呢？

我用的测试文件大小为1GB；内存1GB，CPU双核3.39G，硬盘STAT2接口160G

两个原代码如下：

// Float2.cpp
#include <stdio.h>
#include <time.h>

//#define _MY_DBG_

#ifdef _MY_DBG_
#define print(x) printf x;
#else
#define print(x) ;
#endif

#define DATA_PRE_COUNT (1024*4+3)


int main(int argc, char * argv[])
{
FILE * fp = NULL;
register float fMaxT;
register float fMin=0.0;
register float fMax=0.0;
float buf[DATA_PRE_COUNT];
int offset=0;
int baseOffset=0;

fp = fopen(argv[1], "rb");

clock_t c_start = clock();

int readlen = fread(buf, sizeof(float), DATA_PRE_COUNT, fp);
while (!feof(fp))
{
for (int i=0;i<DATA_PRE_COUNT-3;i++)
{
fMaxT = buf[i]*buf[i+1]*buf[i+2]*buf[i+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+i;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[i],buf[i+1],buf[i+2],buf[i+3],fMax/1e12f));
}
}
baseOffset += DATA_PRE_COUNT-3;
buf[0] = buf[DATA_PRE_COUNT-3];
buf[1] = buf[DATA_PRE_COUNT-2];
buf[2] = buf[DATA_PRE_COUNT-1];
readlen = fread(buf+3, sizeof(float), DATA_PRE_COUNT-3, fp);
if (readlen != (DATA_PRE_COUNT-3))
{
int cnt = readlen-3;
if (cnt < 0)
{
break;
}
for (int j=0;j<cnt;j++)
{
fMaxT = buf[j]*buf[j+1]*buf[j+2]*buf[j+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+j;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[j],buf[j+1],buf[j+2],buf[j+3],fMax/1e12f));
}
}
break;
}
}
clock_t c_end = clock();

printf("乘积=%f 首数的序号=%d\n", fMax, offset);
printf("clocks is %d\n", c_end-c_start);
return 0;
}


// Float3.cpp
#include <stdio.h>
#include <time.h>

//#define _MY_DBG_

#ifdef _MY_DBG_
#define print(x) printf x;
#else
#define print(x) ;
#endif

#define DATA_PRE_COUNT (1024*4+3)


int main(int argc, char * argv[])
{
FILE * fp = NULL;
register float fMaxT;
register float fMin=0.0;
register float fMax=0.0;
float buf[DATA_PRE_COUNT];
int offset=0;
int baseOffset=0;

fp = fopen(argv[1], "rb");

clock_t c_start = clock();

int readlen = fread(buf, sizeof(float), DATA_PRE_COUNT, fp);
while (!feof(fp))
{
/* for (int i=0;i<DATA_PRE_COUNT-3;i++)
{
fMaxT = buf[i]*buf[i+1]*buf[i+2]*buf[i+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+i;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[i],buf[i+1],buf[i+2],buf[i+3],fMax/1e12f));
}
}
baseOffset += DATA_PRE_COUNT-3;
buf[0] = buf[DATA_PRE_COUNT-3];
buf[1] = buf[DATA_PRE_COUNT-2];
buf[2] = buf[DATA_PRE_COUNT-1];
*/
readlen = fread(buf+3, sizeof(float), DATA_PRE_COUNT-3, fp);
/* if (readlen != (DATA_PRE_COUNT-3))
{
int cnt = readlen-3;
if (cnt < 0)
{
break;
}
for (int j=0;j<cnt;j++)
{
fMaxT = buf[j]*buf[j+1]*buf[j+2]*buf[j+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+j;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[j],buf[j+1],buf[j+2],buf[j+3],fMax/1e12f));
}
}
break;
}
*/
}
clock_t c_end = clock();

printf("乘积=%f 首数的序号=%d\n", fMax, offset);
printf("clocks is %d\n", c_end-c_start);
return 0;
}

Float2 data1.dat
乘积=9861133242466304.000000 首数的序号=110335127
clocks is 25812

Float3 data1.dat
乘积=0.000000 首数的序号=0
clocks is 26687

Float2 data1.dat
乘积=9861133242466304.000000 首数的序号=110335127
clocks is 26828

Float3 data1.dat
乘积=0.000000 首数的序号=0
clocks is 27140

[yyc520]

但如果数据少结果就好多了

我用的测试文件大小为64M；内存1GB，CPU双核3.39G，硬盘STAT2接口160G

Float2 data1.dat
乘积=9664972422381568.000000 首数的序号=13430593
clocks is 156
Float3 data1.dat
乘积=0.000000 首数的序号=0
clocks is 78

看样子是要对文件读取进行优化了。

[orlando1108]

看晕了！！！！

[对不起淘淘的失败爸爸]

个人认为：intel测试时会把io的开销不考虑在内，只测试计算时间，因为它提供的代码框架很明显有此意图在内。

By the way， 你的内存load 系统以后，剩余的容量不够contain整个数据，所以会在io时频繁页面和磁盘发生交换，也影响了读数据的性能，尝试小一点的数据说不定测试数据会好一些

[yyc520]

可不是有明确表示，是从程序启动到运行结束所使用的时间么。只有第三道题好像说了要计算TICKS