C++ GetOpenFileName()的问题
我对以下代码有问题:C++ GetOpenFileName()的问题,c++,winapi,visual-c++,C++,Winapi,Visual C++,我对以下代码有问题: HANDLE hFile; DWORD bytesRead; OPENFILENAME ofn; DWORD problem; WCHAR title[260]; ZeroMemory(&ofn, sizeof(ofn)); ofn.lStructSize = sizeof(OPENFILENAME); ofn.hwndOwner = hWnd; ofn.lpstrFile = (LPWSTR)title; ofn.nMaxFile = sizeof(title)
HANDLE hFile;
DWORD bytesRead;
OPENFILENAME ofn;
DWORD problem;
WCHAR title[260];
ZeroMemory(&ofn, sizeof(ofn));
ofn.lStructSize = sizeof(OPENFILENAME);
ofn.hwndOwner = hWnd;
ofn.lpstrFile = (LPWSTR)title;
ofn.nMaxFile = sizeof(title);
ofn.lpstrFilter = TEXT("All files(*.*)\0*.*\0");
ofn.nFilterIndex = 1;
ofn.lpstrInitialDir = NULL;
ofn.Flags = OFN_PATHMUSTEXIST | OFN_FILEMUSTEXIST;
if (GetOpenFileName(&ofn) == false)
{
problem = CommDlgExtendedError();
return false;
}
从GetOpenFileName,它只需转到
problem=commdlgextenderror()
而不显示对话框。您必须为lpstrFile
结构成员分配内存,并将nMaxFile
设置为其大小。此外,缓冲区的第一个字符应设置为\0
,以防止文件名初始化。MSDN示例:
// Initialize OPENFILENAME
ZeroMemory(&ofn, sizeof(ofn));
ofn.lStructSize = sizeof(ofn);
ofn.hwndOwner = hwnd;
ofn.lpstrFile = szFile;
// Set lpstrFile[0] to '\0' so that GetOpenFileName does not
// use the contents of szFile to initialize itself.
ofn.lpstrFile[0] = '\0';
您必须为
lpstrFile
struct成员分配内存,并将nMaxFile
设置为其大小。此外,缓冲区的第一个字符应设置为\0
,以防止文件名初始化。MSDN示例:
// Initialize OPENFILENAME
ZeroMemory(&ofn, sizeof(ofn));
ofn.lStructSize = sizeof(ofn);
ofn.hwndOwner = hwnd;
ofn.lpstrFile = szFile;
// Set lpstrFile[0] to '\0' so that GetOpenFileName does not
// use the contents of szFile to initialize itself.
ofn.lpstrFile[0] = '\0';
…它究竟是什么问题呢?顺便说一句:在这里使用C风格的强制转换只是自找麻烦。它并没有显示一个包含文件列表的窗口。很多事情都很清楚。我的问题是,
问题
变量(或用于进一步指定错误的任何变量)中有什么……它到底是什么问题?顺便说一句:在这里使用C风格的强制转换只是自找麻烦。它并没有显示一个包含文件列表的窗口。很多事情都很清楚。问题
变量(或用于进一步指定错误的任何变量)中有什么是我的问题。