GetOpenFileName触发断点

Question

我尝试这样调用GetOpenFileName：

int main(int argc, char* argv[])
{

OPENFILENAME ofn;       // common dialog box structure
wchar_t szFile[260];       // buffer for file name
HWND hwnd;              // owner window
HANDLE hf;              // file handle
ZeroMemory(&ofn, sizeof(ofn));
ofn.lStructSize = sizeof(ofn);
wchar_t title[500];  // to hold title
GetConsoleTitle( title, 500 );
HWND hwndConsole = FindWindow( NULL, title );
ofn.hwndOwner = hwndConsole;
ofn.lpstrFile = szFile;
ofn.lpstrFile[0] = '\0';
ofn.nMaxFile = sizeof(szFile);
ofn.lpstrFilter = L"All\0*.*\0Text\0*.TXT\0";
ofn.nFilterIndex = 1;
ofn.lpstrFileTitle = NULL;
ofn.nMaxFileTitle = 0;
ofn.lpstrInitialDir = NULL;
ofn.Flags = OFN_PATHMUSTEXIST | OFN_FILEMUSTEXIST;

// Display the Open dialog box. 

if (GetOpenFileName(&ofn)==TRUE) 
    hf = CreateFile(ofn.lpstrFile, 
                    GENERIC_READ,
                    0,
                    (LPSECURITY_ATTRIBUTES) NULL,
                    OPEN_EXISTING,
                    FILE_ATTRIBUTE_NORMAL,
                    (HANDLE) NULL);

程序在"if (GetOpenFileName(&ofn)==TRUE)“处停止(消息: example.exe已触发断点(不是我放置的断点))。当我中断时，我收到一条消息，表明没有可用的源。如果我没有中断，只需按下continue，对话框就会弹出，并按预期工作。我做错了什么？我只是注意到它在发布模式下工作没有问题...

Answer 1

一个可能的问题是：ofn.nMaxFile应该是字符的数量，而不是缓冲区的字节大小。试着这样做：

ofn.nMaxFile = sizeof(szFile) / sizeof(wchar_t);