绝望了，INTEL编程大赛第一题。

2007-07-03 05:11:11 637 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

工业数据采集实战：用open62541库读写PLC的OPC UA节点（避坑指南）本文详细介绍了如何使用open62541库实现工业PLC的OPC UA节点数据采集，涵盖环境搭建、节点读写优化及生产环境故障排查。通过订阅模式和多节点读写技术，显著提升数据采集效率与稳定性，适用于智能制造和工业物联网场景。阅读详情

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

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