控件未在Win32 LPDLGTEMPLATE中添加

时间:2014-09-15 19:58:53

标签: windows winapi

在我的Win32应用程序中,我注册了一个名为" AX的自定义类。"

.rc文件中定义的以下对话框资源按预期工作:

IDD_DIALOG1 DIALOGEX 0, 0, 457, 219
STYLE DS_SETFONT | DS_MODALFRAME | DS_FIXEDSYS | WS_POPUP | WS_CAPTION | WS_SYSMENU | WS_THICKFRAME
CAPTION "Dialog"
FONT 8, "MS Shell Dlg", 400, 0, 0x1
BEGIN
    CONTROL "{8856F961-340A-11D0-A96B-00C04FD705A2}", IDC_EXPLORER1, "AX", WS_CHILD | WS_VISIBLE, 0, 0, 500, 400
END

但是,在我的情况下,我不能使用任何.rc文件。基本上,我需要在内存中构建DLGTEMPLATE并调用DialogBoxIndirect。这是我的相关代码:

lpdt->style = DS_MODALFRAME | DS_FIXEDSYS | WS_POPUP | WS_CAPTION | WS_SYSMENU | WS_THICKFRAME | WS_SYSMENU /*| DS_NOFAILCREATE*/;
lpdt->cdit = 1;         // Number of controls
lpdt->x = 10;  lpdt->y = 10;
lpdt->cx = 500; lpdt->cy = 400;

LPWORD lpw = (LPWORD)(lpdt + 1);
*lpw++ = 0;             // No menu
*lpw++ = 0;             // Predefined dialog box class (by default)

LPWSTR lpwsz = (LPWSTR)lpw;
int nchar = 1 + MultiByteToWideChar(CP_ACP, 0, "My Dialog", -1, lpwsz, 50);
lpw += nchar;

////-----------------------
//// Define the WebBrowser Control.
////-----------------------
lpw = lpwAlign(lpw); 
LPDLGITEMTEMPLATE lpdit = (LPDLGITEMTEMPLATE)lpw;
lpdit->x  = 0; lpdit->y  = 0;
lpdit->cx = 300; lpdit->cy = 300;
lpdit->id = IDC_EXPLORER1; 
lpdit->style = WS_CHILD | WS_VISIBLE;

// Class for the activeXControl
lpw = (LPWORD)(lpdit + 1);
*lpw++ = 0xFFFF;
*lpw++ = atom; // Here, atom was obtained from RegisterClassEx() for class name "AX"

lpw = (LPWORD) (lpdit + 1);
lpwsz = (LPWSTR) lpw;
nchar = 1 + MultiByteToWideChar (CP_ACP, 0, "{8856F961-340A-11D0-A96B-00C04FD705A2}", -1, lpwsz, 2048); 
lpw   += nchar;
*lpw++ = 0; // no creation data

GlobalUnlock(hgbl);
LRESULT ret = DialogBoxIndirect(hinst, (LPDLGTEMPLATE)hgbl, hwndOwner, DialogProc);
GlobalFree(hgbl);

if (ret == -1) {
    int err = ::GetLastError();
    ReportError(err);
}

DialogBoxIndirect的返回码是-1。但是,GetLastError()返回0. Strange。

如果我在对话框样式中指定DS_NOFAILCREATE,则会创建对话框。但是,我需要的控制缺失。

请帮我识别代码中的问题。问候。

2 个答案:

答案 0 :(得分:1)

看起来您的错误可能就在这里:

// Class for the activeXControl
lpw = (LPWORD)(lpdit + 1);
*lpw++ = 0xFFFF;
*lpw++ = atom; // Here, atom was obtained from RegisterClassEx() for class name "AX"

lpw = (LPWORD) (lpdit + 1);   // <--- ERROR HERE
lpwsz = (LPWSTR) lpw;

您再次将lpw重新分配给lpdit + 1,这意味着您可以在类原子上编写控件标题文本。看起来像是一个糟糕的复制+粘贴工作。

答案 1 :(得分:0)

让编译器创建对话框模板。只需使用二进制编辑器,只需复制序列并将其定义为char数组,或其他任何内容。

这就是我这样做的方式,它避免了错误。