我想测量/优化应用程序的“冷启动”启动性能，并且在没有实际重新启动的情况下很难做到这一点，这显然不是理想的解决方案。
是否有一种方法可以使整个系统的文件缓存无效，从而使映射的页面访问实际上导致磁盘访问，从而可以测量程序启动所花费的时间？
我非常需要FSCTL_DISMOUNT_VOLUME的功能，但要占用系统空间。
至少在Windows 7上，似乎试图打开卷句柄无 FILE_SHARE_WRITE共享权限导致文件系统缓存无效，即使创建失败。
因此，我编写了一个为此简单地调用CreateFile的程序。
程序的Base64编码*：
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// Usage: ClearCache C: D:
#include <tchar.h>
#include <stdio.h>
#include <windows.h>
int _tmain(int argc, LPTSTR argv[]) {
LPCTSTR DOS_PREFIX = _T("\\\\.\\");
for (int i = 1; i < argc; i++) {
LPTSTR arg = argv[i];
LPTSTR path = (LPTSTR)calloc(
_tcslen(arg) + _tcslen(DOS_PREFIX) + 1, sizeof(*arg));
__try {
if (_istalpha(arg[0]) && arg[1] == _T(':') &&
(arg[2] == _T('\0') ||
arg[2] == _T('\\') && arg[3] == _T('\0')))
{ _tcscat(path, DOS_PREFIX); }
_tcscat(path, arg);
HANDLE hFile = CreateFile(path,
FILE_READ_DATA, FILE_SHARE_READ, NULL, OPEN_EXISTING, 0, NULL);
if (hFile != INVALID_HANDLE_VALUE) { CloseHandle(hFile); }
else {
DWORD le = GetLastError();
if (le != ERROR_SHARING_VIOLATION && le != ERROR_ACCESS_DENIED)
{
_ftprintf(stderr, _T("Error %d clearing %s\n"), le, argv[i]);
return le;
}
}
} __finally { free(path); }
}
return 0;
}
*只是为了好玩，看看是否可以通过反汇编来弄清楚可执行文件的功能。这不是您的典型可执行文件。 ：）
我编写了一个简单的命令行实用程序来做到这一点：FlushFileCache
它依赖于未记录的NtSetSystemInformation函数，并且也可以刷新其他各种内存池。
戴维说。创建一个大文件，但是需要多少GB，并且每次要重置文件缓存时，都请复制该文件。然后确保删除旧文件。
因此，创建BIGFILE1.DAT，将其复制到BIGFILE2.DAT，然后删除BIGFILE1.DAT（将其从磁盘和缓存中删除）。下次，只需逆转该过程即可。
附录：
嗯，另一种选择是获取已映射的文件，并将其复制到新文件，删除旧文件，然后将新文件重命名为旧文件。缓存由文件支持。如果文件“消失”，缓存也将消失。
如果可以识别这些文件，并且系统/其他正在运行的程序不共享这些文件，则该脚本应该易于编写，并且理想情况下，运行速度要比复制6G的文件快。
此解决方案效果很好：Clear file cache to repeat performance testing
更具体地说，我正在这样做：
// Open with FILE_FLAG_NO_BUFFERING
auto hFile = CreateFile(path.c_str(),
GENERIC_READ,
FILE_SHARE_READ,
nullptr,
OPEN_EXISTING,
FILE_FLAG_NO_BUFFERING,
nullptr);
/// Check
if (hFile == INVALID_HANDLE_VALUE){
//_tprintf(TEXT("Terminal failure: unable to open file \"%s\" for read.\n"), argv[1]);
cout << "error" << endl;
return;
}
// Close
CloseHandle(hFile);
// Now open file with regular C++ API, and caching disabled
ifstream file(path, ios::binary | ios::ate);
您可以使用VM并在VM引导后立即进行快照。从快照恢复将比重新启动更快。