请教高手指点

#include <stdio.h>
#include <AT89x051.H>
#define liekey (~P0)&0X0F
#define rowkey (~P0)&0XF0
sfr P0 = 0x80;
sfr P2 = 0xA0;
void Initial();
int Data_In();
int Comput();
bit KeyIn(void);  //如果有键按下返回 1，健值存在 KEYdata
int OP1,OP2,OP,sta;

char KeyData,Buff,Resut;
char DisBuffer[8];
char KeyData=0;
char KeyTab[]={0x7e,0x7d,0x7b,0x0be,0x0bd,0x0bb,0x0de,0x0dd,0x0db,0x0ee,0x77,0x0b7,0x0d7,0x0e7,0x0eb,0x0ed};//键盘的16个键
char Seg_Tab[]={0x3F,0x06,0x5B,0x4F,0x66,0x6D,0x7D,0x07,0x7F,0x6F,0x79};//数字的0~9 + - */ 对应 ASCD  "E"对应错误    
      

void main ()
{  bit flag=0; 
   Initial();
  while (1)

   {  flag=KeyIn();
   if( flag)
   {
            DisBuffer[1]=KeyData;
           //  Display();

      }
     else
   {
   }

        
 
 }
}

  bit KeyIn()
{   int i;
    char lie,kcode,row; //lie为列对应的低位数据的取反 row为行对应的高位数据取反
  bit keyin=0,last,temp;
     lie=liekey;    
     row=rowkey;
    if(lie!=0)
    {           //若低位出现非0的情况则为有键按下
       kcode=~(lie|row);  //将按键所对应的二进制数据送给kcode
     for(i=0;KeyTab!=kcode&&i<16;i++) 
           KeyData=i+1;
           keyin=1;
     }
     last=keyin;
 
    keyin=keyin&temp;
    temp=~last;
   return keyin;
 }
 void Initial()
 { OP1=0;
   OP2=0;
   sta=0;
   TH0=0xd8;      // ;定时时间为10ms
   TL0=0xf0;
   ET0=1;
   EA=1;
   TR0=1;
   DisBuffer[0]=0x3f;
   DisBuffer[1]=0x3f;
   DisBuffer[2]=0x3f;
    DisBuffer[3]=0x3f;

已经解决