windows编程基础有关问题GetOpenFileName的用法

windows编程基础问题GetOpenFileName的用法
我欲获取路径和文件名，用的下面的语句，

C/C++ code

if(GetOpenFileName(&ofn)==TRUE)  MessageBox (hWnd,ofn.lpstrFile,L"文件名",MB_OK);

打开文件对话框是能打开的，但是ofn.lpstrFile结果在BOX上显示是空白，查看变量ofn.lpstrFile值就是0x000000是什么原因？
关于ofn我是这样定义的

C/C++ code

      OPENFILENAME ofn;            static TCHAR szFilter[] =TEXT ("All Files (*.*)\0*.*\0\0") ;       ofn.lStructSize=sizeof (OPENFILENAME) ;       ofn.hwndOwner             = NULL ;       ofn.hInstance             = NULL ;       ofn.lpstrFilter           = szFilter ;       ofn.lpstrCustomFilter  = NULL ;       ofn.nMaxCustFilter     = 0 ;       ofn.nFilterIndex       = 0 ;       ofn.lpstrFile           = NULL;                ofn.nMaxFile              = MAX_PATH ;       ofn.lpstrFileTitle       = NULL ;                ofn.nMaxFileTitle        = MAX_PATH ;       ofn.lpstrInitialDir      = NULL ;       ofn.lpstrTitle            = NULL ;       ofn.Flags                 = OFN_PATHMUSTEXIST | OFN_FILEMUSTEXIST ;      ofn.nFileOffset           = 0 ;       ofn.nFileExtension       = 0 ;       ofn.lpstrDefExt           = NULL;       ofn.lCustData             = 0L ;       ofn.lpfnHook              = NULL ;      ofn.lpTemplateName       = NULL ; 

[解决办法]
http://blog.csdn.net/caimouse/article/details/1958035

查看更多 下一篇