m_Operation[i]存储的是运算符，1+ 2- 3* 4/，   m_number[i]存储的是操作数，代码如下，如何实现连乘连除功能？
void CCalculatorDlg::cal()
{
    for(int i=1;i<15;i++)
        switch(m_Operation[i])
    {
    case 3:
        m_number[i]=m_number[i+1]=m_number[i]*m_number[i+1];
        break;
    case 4:
        m_number[i]=m_number[i+1]=(double)m_number[i]/m_number[i+1];
        break;
    }
    m_result=m_number[1];
    for(i=1;i<15;i++)
        if(m_Operation[i]==1)
            m_result=m_result+m_number[i+1];
        else if (m_Operation[i]==2)
            m_result=m_result-m_number[i+1];
            UpdateData(FALSE);
}

解决了：
void CCalculatorDlg::cal()
{
    int curi=0,flagg=0,flagg1=0;
    double nown=1,kret[15];
    for(int i=1;i<15;i++)
    {
        switch(m_Operation[i])
        {
        case 3:
            if (flagg==0)//第一次
            {
                nown=m_number[i]*m_number[i+1];
                flagg=1;
            }
            else
            {
                nown*=m_number[i+1];
            }
            break;
        case 4:
            if (flagg==0)//第一次
            {
                nown=(double)m_number[i]/m_number[i+1];
                flagg=1;
            }
            else
            {
                nown=(double)nown/m_number[i+1];
            }
            break;
        case 1:
        case 2:
            if (flagg==1)
            {
                kret[curi++]=nown;
            }
        case 999://运算结束
            if (m_Operation[i-1]==3 || m_Operation[i-1]==4)
            {//最后一个操作符是乘或除的话
                kret[curi++]=nown;
                flagg1=1;
            }
            break;
        }
        if (flagg1==1)
        {
            break;
        }
    }
    int nowq=0;
    if (m_Operation[1]==3 || m_Operation[1]==4)
    {
        m_result=kret[nowq++];
    }
    else
    {
        m_result=m_number[1];
    }
    for(i=1;i<15;i++)
        if(m_Operation[i]==1)
            if (m_Operation[i+1]==3 || m_Operation[i+1]==4)
                m_result=m_result+kret[nowq++];
            else
                m_result=m_result+m_number[i+1];
            else if (m_Operation[i]==2)
                if (m_Operation[i+1]==3 || m_Operation[i+1]==4)
                    m_result=m_result-kret[nowq++];
                else
                    m_result=m_result-m_number[i+1];
            UpdateData(FALSE);


}