实践中遇到有符号十六进制数需要转化成十进制数，编程如下，如有疏漏，恳请指出。

int hextode(char* hex)
{
    if(hex ==NULL)
        return 0;
    char binary[17]={0};//放二进制字符串
    char* hexstr="006C";//一个16进制字符串
    //Convert strings to a long-integer value.
    long i32=strtol(hexstr,NULL,16);//读取16进制的值
    cout<<i32<<endl;

    int toint = 0;
    int ratio = 1;
    if(hexstr[0]>'8')
    {
       cout<<"此数为负"<<endl;
       //Converts an integer to a string
       itoa(i32,binary,2);/*输出为二进制*/
       for(int i=0;i<16;i++) 
          cout<<binary[i];
       cout<<endl;
       cout<<' ';
       for(int i=1;i<16;i++) //除第一位外取反
            {
             binary[i] ='0'+!(binary[i]-'0');
             cout<<binary[i];
       }
       cout<<endl;
       int nTakeover = 0;
       bool isoverflow = false;
       for(int i=15;i>=0;i--) //二进制加1
       {
          int nsum = binary[i]-'0'+nTakeover; //每位加上进位
          if(i==15)
              nsum++;
          if(nsum==2)
              {
                  if(i == 0)
                      isoverflow = true;
                  else
                  {
                  nsum-=2;
                  nTakeover = 1;
                  binary[i] = '0'+nsum;
                  }
              }
          else
          {
          binary[i] = '0'+nsum;
          break;
          }
       } 
       for(int i=0;i<16;i++) //取反加一后
            {
             cout<<binary[i];
       }
       cout<<endl;
       for(int j=15;j>0;--j)
       {
         toint = toint+ (binary[j]-'0')*ratio;
         ratio=ratio*2;
       }    
       toint = toint*(-1);
       cout<<toint<<endl;
       return toint;
    }
    else
    {
       cout<<"此数为正"<<endl;
       toint= i32;
       cout<<toint<<endl;
       return toint;
    }
}