c++ windows下读取大文件（内存映射）
关于内存映射的基本知识以及一些函数的原型说明，参考博客：http://blog.csdn.net/wcyoot/article/details/7363393
下面是我对于读取一个104M文件大小，使用内存映射和传统的读文件方法进行了时间上的对比，两种方法都从该文讲读取300000*34个浮点数。
测试环境：vs2010
1 #include<iostream> 2 #include <cstringt.h> 3 #include<vector> 4 #include<windows.h> 5 #include <ctime> 6 using namespace std; 7 8 9 10 11 int main() 12 { 13 clock_t start, finish; 14 //------------------------------------------------------内存映射方法; 15 start = clock(); 16 //创建或打开文件内核对象; 17 HANDLE fileH = CreateFile("ldFeature.txt", 18 GENERIC_READ|GENERIC_WRITE, 19 FILE_SHARE_READ, 20 NULL, 21 OPEN_EXISTING, 22 FILE_ATTRIBUTE_NORMAL, 23 NULL); 24 if(fileH == INVALID_HANDLE_VALUE) 25 { 26 cout<<"error in CreateFile"<<endl; 27 return -1; 28 } 29 30 //创建一个文件映射内核对象; 31 HANDLE mapFileH = CreateFileMapping( fileH, 32 NULL, 33 PAGE_READWRITE, 34 0, 35 0, 36 "Resource " ); 37 if(mapFileH == NULL) 38 { 39 cout<<"error in CreateFileMapping"<<endl; 40 return -1; 41 } 42 43 //将文件数据映射到进程的地址空间; 44 char * mapH = (char *)MapViewOfFile( mapFileH, 45 FILE_MAP_ALL_ACCESS, 46 0, 47 0, 48 0); 49 if(mapH == NULL) 50 { 51 cout<<"error in MapViewOfFile"<<endl; 52 return -1; 53 } 54 55 //读取数据; 56 char *buf = mapH; 57 double k; 58 int times = 300000*34; 59 for(int i = 1; i <= times; i++) 60 { 61 k = atof(buf); 62 buf = strstr(buf+9," "); 63 //cout<<k<<endl; 64 } 65 //关闭句柄; 66 UnmapViewOfFile(mapH); 67 CloseHandle(mapFileH); 68 CloseHandle(fileH); 69 70 finish = clock(); 71 //打印耗时; 72 cout<<"time:"<<(double)(finish - start)/ CLOCKS_PER_SEC<<endl; 73 74 //------------------------------------------------------传统方法; 75 start = clock(); 76 77 FILE *fp = fopen("ldFeature.txt","r"); 78 if(fp == NULL) 79 { 80 cout<<"error in fopen"<<endl; 81 return -1; 82 } 83 for(int i = 1 ;i <= times; i++) 84 { 85 fscanf(fp,"%lf",&k); 86 } 87 fclose(fp); 88 89 finish = clock(); 90 cout<<"time:"<<(double)(finish - start)/ CLOCKS_PER_SEC<<endl; 91 92 93 return 0; 94 }
结果显示，传统方法耗时是内存映射方法的2倍左右。实际上文件越大，使用内存映射方法的优势越明显：
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