-
#define uint unsigned int
#define uchar unsigned char
sbit k1=P2^0;
sbit k2=P2^1;
uint code table[]={0X00,0X80,0X10,0X90,0X20,0XA0,0X30,0XB0,0X40,0XC0 };
void delay(uint z)
{
uint a,b;
for(a=z;a>0;a--)
{
for(b=z;b>0;b--);
}
}
void main()
{
int c=99,t,shi,ge;
for(t=99;t>0;t--)
{
shi=c/10;
ge=c;
k1=0,k2=1;
P1=table[shi ];
delay(300);
P1=0XFF;
k2=0,k1=1;
P1=table[ge];
delay(200);
P1=0XFF;
k1=0,k2=0;
c--;
if(c ==0)
c=99;
}
}
- answer:
-
It must be different at the same time, because you showed it as ten and then one.
In fact, the so-called simultaneous display means that if the cycle scanning time interval is short, it will naturally appear to be simultaneous.
What you have to do is to display the tens digit and then the ones digit while displaying the delay , instead of displaying one and then the other.
The best way is to put it in the interrupt to ensure the display effect.
bit flag;
int shi ,ge;
These three are external variables
and it should be OK to call this function during the delay process. void shown()
{
if (flag==0)
{
P1=table[shi];k2=1,k1=0;
}
else
{
P1=table[ge];k2=0,k1=1;
}
flag=~flag;
}
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