win32 中 BS_AUTOCHECKBOX 请教

Question

加法器小改进成支持加减运算，遇到一个问题， 在 WM_COMMAND 的处理中实现不了改变运算符号，不知道是 BS_AUTOCHECKBOX 的用法错误了，还是其他原因，请大家指教，代码如下：
#include <windows.h>
#include <tchar.h>
#include <stdlib.h>
#include <stdio.h>
HWND hwin;
HWND hedt1;
HWND hedt2;
HWND hbtn;
HWND hresult;
HWND hsta1;
HWND hsta2;
HWND hbtn_op;
int op=0;LRESULT CALLBACK MyWinProc(HWND hwnd,UINT uMsg,WPARAM wParam, LPARAM lParam)
{
    int r;
    double x;
    double y;
    char buf[30];
    WORD w;
    HWND htmp;
LRESULT res;    switch (uMsg)
    {
    case WM_SYSCOMMAND:
        w=(UINT)wParam;
        if(w==SC_MINIMIZE || w==SC_MAXIMIZE )
        {
            return 0;
        }
    case WM_COMMAND:
        htmp =(HWND)lParam;
if( htmp == hbtn_op)  //改变运算符号
{
res = SendMessage( hbtn_op,BM_GETSTATE,0,0);
if( res == BST_CHECKED )
{
op =0;
SetWindowText(hsta1,"+");
}
if( res == BST_UNCHECKED )
{
op =1;
SetWindowText(hsta1,"-");
}
return 0;
}
        if( htmp == hbtn )
        {
            memset(buf,0,30);
            GetWindowText(hedt1,buf,29);
x=atof(buf);
            if( x==0 )
            {
                MessageBox(hwnd,"输入的数据有误！","错误",MB_OK);
                return 0;
            }
            memset(buf,0,30);
            GetWindowText(hedt2,buf,29);
            y=atof(buf);
            if( y==0 )
            {
                MessageBox(hwnd,"输入的数据有误！","错误",MB_OK);
                return 0;
            }
            switch (op)
{
case 0:
x+=y;
break;
case 1:
x=y;
break;
}
            sprintf(buf,"%g",x);
            SetWindowText(hresult,buf);
            return 0;        }        if( htmp==hedt1||htmp==hedt2 )//有输入清空结果
        {
            SetWindowText(hresult,"");
            return 0;
        }
        break;
    case WM_CLOSE:
        r=MessageBox(hwnd,"是否退出","退出",MB_YESNO);
        if(r==IDYES)
        {
            DestroyWindow(hwnd);
        }
        return 0;
    case WM_DESTROY:
        PostQuitMessage(0);
        return 0;
    }
    return DefWindowProc(hwnd,uMsg,wParam,lParam);
}
int WINAPI WinMain( HINSTANCE hins, HINSTANCE  hpre,LPSTR cmd,int showtype)
{
    WNDCLASSEX wnd;
    MSG msg;
    BOOL bRet;     wnd.cbClsExtra=0;
    wnd.cbSize=sizeof(wnd);
    wnd.cbWndExtra=0;
    wnd.hbrBackground=(HBRUSH)(COLOR_BTNFACE+1);
    wnd.hCursor=0;
    wnd.hIcon=0;
    wnd.hIconSm=0;
    wnd.hInstance=GetModuleHandle(0);
    wnd.lpfnWndProc= MyWinProc;
    wnd.lpszClassName="mywin";
    wnd.lpszMenuName=0;
    wnd.style=CS_HREDRAW|CS_VREDRAW;    RegisterClassEx(&wnd);
    hwin=CreateWindowEx(WS_EX_CLIENTEDGE,"mywin","加法器",WS_OVERLAPPEDWINDOW|WS_VISIBLE,
                        0,0,400,300,0,0,hins,0);
    hedt1=CreateWindowEx(0,_T("EDIT"),_T(""),WS_CHILD|WS_VISIBLE,
                        50,60,100,20,hwin,0,hins,0);
    hedt2=CreateWindowEx(0,_T("EDIT"),_T(""),WS_CHILD|WS_VISIBLE,
                        200,60,100,20,hwin,0,hins,0);
    hbtn=CreateWindowEx(0,_T("BUTTON"),_T("计算结果"),WS_CHILD|WS_VISIBLE,
                        120,90,100,30,hwin,0,hins,0);
    hsta1=CreateWindowEx(0,_T("STATIC"),_T("+"),WS_CHILD|WS_VISIBLE,
                        170,60,10,20,hwin,0,hins,0);
    hresult=CreateWindowEx(0,_T("STATIC"),_T(""),WS_CHILD|WS_VISIBLE,
                        150,140,200,20,hwin,0,hins,0);
    hsta2=CreateWindowEx(0,_T("STATIC"),_T("结果是："),WS_CHILD|WS_VISIBLE,
                        50,140,70,30,hwin,0,hins,0);
hbtn_op=CreateWindowEx(0,_T("BUTTON"),_T("加/减 选择"),WS_CHILD|WS_VISIBLE | BS_AUTOCHECKBOX ,
                        120,1,120,30,hwin,0,hins,0);
SendMessage( hbtn_op,BM_SETCHECK,BST_CHECKED,0);
    while( ( bRet=GetMessage(&msg,0,0,0) ) !=0 )
    {
        if( bRet==-1)
        {
            return -1;
        }
        else
        {
            TranslateMessage(&msg);
            DispatchMessage(&msg);
        }
    }
    return 0;
}

Answer 1

case 1:
   x=y;
   break;
你要减法，怎么是 x=y  ？

Answer 2

BS_AUTOCHECKBOX 这个和窗体风格有什么关系啊？是不是要重绘下界面啊

Answer 3

res = SendMessage( hbtn_op,BM_GETSTATE,0,0);
-->
res = SendMessage( hbtn_op,BM_GETCHECK,0,0);switch (op)
            {
            case 0:
                x+=y;
                break;
            case 1:
                x=y;
                break;
            }
中的case 1:修改为
case 1:
 x -= y;
 break;