我有一个SSD，正在尝试使用它来模拟程序的I / O性能，但是，从程序中计算出的IOPS比IOMeter快得多。
我的SSD是IOMeter的PLEXTOR PX-128M3S，它的最大512B随机读取IOPS约为94k(队列深度为32)。
但是我的程序(32个Windows线程)可以达到大约500k 512B IOPS，大约是IOMeter的5倍!我进行了数据验证，但是在数据获取中没有发现任何错误。这是因为我的数据按顺序提取吗？
我粘贴了我的代码belwo(主要是从文件中获取512B并释放它；我确实使用4bytes(一个int值)来验证程序逻辑，但没有发现问题)，有人可以帮助我找出我的错误地方吗？
提前非常感谢!
#include <stdio.h>
#include <Windows.h>
//Global variables
long completeIOs = 0;
long completeBytes = 0;
int threadCount = 32;
unsigned long long length = 1073741824; //4G test file
int interval = 1024;
int resultArrayLen = 320000;
int *result = new int[resultArrayLen];
//Method declarison
double GetSecs(void); //Calculate out duration
int InitPool(long long,char*,int); //Initialize test data for testing, if successful, return 1; otherwise, return a non 1 value.
int * FileRead(char * path);
unsigned int DataVerification(int*, int sampleItem); //Verify data fetched from pool
int main()
{
int sampleItem = 0x1;
char * fPath = "G:\\workspace\\4G.bin";
unsigned int invalidIO = 0;
if (InitPool(length,fPath,sampleItem)!= 1)
printf("File write err... \n");
//start do random I/Os from initialized file
double start = GetSecs();
int * fetchResult = FileRead(fPath);
double end = GetSecs();
printf("File read IOPS is %.4f per second.. \n",completeIOs/(end - start));
//start data validation, for 4 bytes fetch only
// invalidIO = DataVerification(fetchResult,sampleItem);
// if (invalidIO !=0)
// {
// printf("Total invalid data fetch IOs are %d", invalidIO);
// }
return 0;
}
int InitPool(long long length, char* path, int sample)
{
printf("Start initializing test data ... \n");
FILE * fp = fopen(path,"wb");
if (fp == NULL)
{
printf("file open err... \n");
exit (-1);
}
else //initialize file for testing
{
fseek(fp,0L,SEEK_SET);
for (int i=0; i<length; i++)
{
fwrite(&sample,sizeof(int),1,fp);
}
fclose(fp);
fp = NULL;
printf("Data initialization is complete...\n");
return 1;
}
}
double GetSecs(void)
{
LARGE_INTEGER frequency;
LARGE_INTEGER start;
if(! QueryPerformanceFrequency(&frequency))
printf("QueryPerformanceFrequency Failed\n");
if(! QueryPerformanceCounter(&start))
printf("QueryPerformanceCounter Failed\n");
return ((double)start.QuadPart/(double)frequency.QuadPart);
}
class input
{
public:
char *path;
int starting;
input (int st, char * filePath):starting(st),path(filePath){}
};
//Workers
DWORD WINAPI FileReadThreadEntry(LPVOID lpThreadParameter)
{
input * in = (input*) lpThreadParameter;
char* path = in->path;
FILE * fp = fopen(path,"rb");
int sPos = in->starting;
// int * result = in->r;
if(fp != NULL)
{
fpos_t pos;
for (int i=0; i<resultArrayLen/threadCount;i++)
{
pos = i * interval;
fsetpos(fp,&pos);
//For 512 bytes fetch each time
unsigned char *c =new unsigned char [512];
if (fread(c,512,1,fp) ==1)
{
InterlockedIncrement(&completeIOs);
delete c;
}
//For 4 bytes fetch each time
/*if (fread(&result[sPos + i],sizeof(int),1,fp) ==1)
{
InterlockedIncrement(&completeIOs);
}*/
else
{
printf("file read err...\n");
exit(-1);
}
}
fclose(fp);
fp = NULL;
}
else
{
printf("File open err... \n");
exit(-1);
}
}
int * FileRead(char * p)
{
printf("Starting reading file ... \n");
HANDLE mWorkThread[256]; //max 256 threads
completeIOs = 0;
int slice = int (resultArrayLen/threadCount);
for(int i = 0; i < threadCount; i++)
{
mWorkThread[i] = CreateThread(
NULL,
0,
FileReadThreadEntry,
(LPVOID)(new input(i*slice,p)),
0,
NULL);
}
WaitForMultipleObjects(threadCount, mWorkThread, TRUE, INFINITE);
printf("File read complete... \n");
return result;
}
unsigned int DataVerification(int* result, int sampleItem)
{
unsigned int invalid = 0;
for (int i=0; i< resultArrayLen/interval;i++)
{
if (result[i]!=sampleItem)
{
invalid ++;
continue;
}
}
return invalid;
}
最佳答案
我没有找到足够详细的信息，但是没有看到任何代码可以将数据刷新到磁盘和/或确保您的读取实际上来自磁盘。事实就是如此，看来您要衡量的主要是操作系统磁盘缓存的性能。尽管磁盘可能会对您要衡量的性能有所贡献，但它可能只是一个很小的贡献者，而其他因素则占主导地位。
由于代码显然是为Windows编写的，因此(例如，一个示例)您可以考虑使用CreateFile打开文件，并在这样做时传递FILE_FLAG_NO_BUFFERING标志。这将(至少大部分情况下)从等式中删除操作系统缓存，并强制每次读取或写入直接处理磁盘本身。
关于c++ - 为什么Windows C++多线程IOPS比IOMeter快得多？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/9810709/