关于#单片机#的问题，如何解决？

为什么两组数码管初值设置不同但是仿真时间数值同步

#include
#define uint unsigned int           
#define uchar unsigned char   
uchar tab2[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};    

uchar t1[1];
uchar t2[1];

uchar sum=61;      //数码管初始值
uchar su=50;
sbit P22=P2^2; //段
sbit P23=P2^3;
sbit P24=P2^4;        

sbit P10=P1^0;  //东西 红 黄 绿
sbit P11=P1^1;
sbit P12=P1^2;

sbit P14=P1^4;  //南北 红 黄 绿
sbit P15=P1^5;
sbit P16=P1^6;

sbit P31=P3^1;   //按键
sbit P30=P3^0;
sbit P32=P3^2;
sbit P33=P3^3;



void delay(uint ms)                 //延时
{
int i,j;
for(i=0;i
    for(j=0;j<120;j++);
}

void smxs( uchar ws, uchar ds )    
{
switch(ws)  
   {
   case 1:  P22=0; P23=0; P24=0;  break;
   case 2:  P22=1; P23=0; P24=0;  break;
   case 3:  P22=0; P23=1; P24=0;  break;
   case 4:  P22=1; P23=1; P24=0;  break;
   case 5:  P22=0; P23=0; P24=1;  break;
   case 6:  P22=1; P23=0; P24=1;  break;
   case 7:  P22=0; P23=1; P24=1;  break;
   case 8:  P22=1; P23=1; P24=1;  break;
   }
P0=tab2[ds]; 
}
void smx( uchar ws, uchar ds )    
{
switch(ws)  
   {
   case 1:  P22=0; P23=0; P24=0;  break;
   case 2:  P22=1; P23=0; P24=0;  break;
   case 3:  P22=0; P23=1; P24=0;  break;
   case 4:  P22=1; P23=1; P24=0;  break;
   case 5:  P22=0; P23=0; P24=1;  break;
   case 6:  P22=1; P23=0; P24=1;  break;
   case 7:  P22=0; P23=1; P24=1;  break;
   case 8:  P22=1; P23=1; P24=1;  break;
   }
P0=tab2[ds]; 
}
void m()
{
    t1[1]=sum/10;
    t1[0]=sum%10;
    t2[1]=su/10;
    t2[0]=su%10;    
    
}    
    

void time_0()
{
TMOD=0x01;
TH0=0X3c;    
TL0=0Xb0;    
TR0=1;
EA=1;
ET0=1;
    
}

void main()
{
 time_0();



    while(1)
    { m();        
      smxs(1,t1[1]);
    delay(1);
    smxs(2,t1[0]);
    delay(1);    
      smx(4,t2[1]);
    delay(1);
    smx(3,t2[0]);
    delay(1);            
    }
        
    
}    
void t0()interrupt 1
{
    static uchar counter=0;
        TH0=0X3c;    
        TL0=0Xb0;    
        counter++;
     if(counter==20)
     { sum--;
     su--;
         counter=0;     
     } 
    
}

略看数组T1T2只声明1个空间是不存在下标1的