我创建了一个软键,想把它放在手机屏幕的左下脚,宽度不能超过手机屏幕宽度的一半,我该怎么写程序?急,谢谢!

[木子月月鱼]

我用下面这段程序,结果软键都显示不出来了,请教高手指点!谢谢!
AEERect qrc;
qrc.x = pim_brew->DeviceInfo.cxScreen-50;
qrc.y = pim_brew->DeviceInfo.cxScreen-20;
qrc.dx = pim_brew->DeviceInfo.cxScreen;
qrc.dy = pim_brew->DeviceInfo.cxScreen;

IMENUCTL_SetRect(pMe->pIExitMenuCtl,&qrc); //设置矩形框
IMENUCTL_AddItem(pMe->pIExitMenuCtl, NULL, NULL, ITEM_SOFTMENU_EXIT, L"Synchron Exit", NULL);
IMENUCTL_SetActive(pMe->pIExitMenuCtl, TRUE); // 激活菜单控件对象
IMENUCTL_Redraw(pMe->pIExitMenuCtl); //刷新菜单控件显示

[sunwhite]

你定义的qrc似乎有问题，你应该检查pim_brew->DeviceInfo.cxScreen-50和pim_brew->DeviceInfo.cxScreen-20是否超出了屏幕显示范围