Please help me with external interrupt and digital tube display (just look at the problem of interrupt trigger)

liwing666

Please help me with external interrupt and digital tube display (just look at the problem of interrupt trigger) [Copy link]

Title: Drive the digital tube to display dynamically, and move from right to left on 8 digital tubes and display the number "2019" cyclically. Use the toggle switch or key as the external interrupt request signal to execute the interrupt service
program:
(1) If the interrupt is entered for the first time, all digital tubes will display 1, and the first light on the right side of the LED will keep
flashing;
(2) If the interrupt is entered for the second time, all digital tubes will display 2, and the first two lights on the right side of the LED will keep
flashing;
...;
(6) If the interrupt is entered for the sixth time, all digital tubes will display 6, and the first six lights on the right side of the LED will keep
flashing;
(7) If the interrupt is entered for the seventh time, the last 8 digits of your student ID will be displayed on the digital tube, and all lights on the LED
will keep flashing.
If there is another interrupt, the next cycle will start from (1).

Code:

#include<reg52.h>
#include<intrins.h> //crol function call header file
#define ui unsigned int //macro definition
#define uc unsigned char
sbit dula=P2^6; //define segment select latch
sbit wela=P2^7; //bit select latch
sbit key=P3^4; //define key
sbit stop=P3^2; //external interrupt 0
uc d;
ui b;
ui i;
ui a;
uc c;
//realize six-digit digital tube scrolling display 2019
uc code table[]={0x5b,0x3f,0x06,0x6f}; //display "2019"
uc code table1[]={0x3e,0xfd,0xfb,0xf7}; //define digital tube 1, 2, 3, 4 digit display
uc code table2[]={0xfd,0xfb,0xf7,0xef}; //Define 2, 3, 4, 5-digit display of digital tube
uc code table3[]={0xfb,0xf7,0xef,0xdf}; //Define 3, 4, 5, 6-digit display of digital tube
uc code table4[]={0xf7,0xef,0xdf,0x3e}; ///Define 4, 5, 6, 1-digit display of digital tube/
uc code table5[]={0xef,0xdf,0x3e,0xfd}; //Define 5, 6, 1, 2-digit display of digital tube
uc code table6[]={0xdf,0x3e,0xfd,0xfb}; //Define 6, 1, 2, 3-digit display of digital tube
void delay(ui xms) //Delay function
{
ui x,y;
for(x=xms;x>0;x--)
for(y=110;y>0;y--);
}
void display() //六位数码管滚动显示2019
{
for(a=650;a>0;a--)
{
for(i=0;i<4;i++){
dula=1;
P0=table[i];
dula=0;
P0=0xff;
wela=1;
P0=table1[i];
wela=0;
delay(1);
}
}

for(a=650;a>0;a--)
{
for(i=0;i<4;i++){
dula=1;
P0=table[i];
dula=0;
P0=0xff;
wela=1;
P0=table2[i];
wela=0;
delay(1);
}
}

for(a=650;a>0;a--)
{
for(i=0;i<4;i++){
dula=1;
P0=table[i];
dula=0;
P0=0xff;
wela=1;
P0=table3[i];
wela=0; delay
(1);
}
}

for(a=650;a>0;a--)
{ for(i = 0;i< 4 ; i
++){ dula =1 ; } for(a=650;a>0;a--) { for ( i=0;i<4;i++){ dula = 1 ; P0 = table [i]; dula =0 ; P0 = 0xff ; wela =1 ; P0=table[i]; dula=0; P0=0xff; wela=1;

































P0=table6[i];
wela=0;
delay(1);
}
}

}void init() //initial function
{
EA=1; //enable total interrupt
EX0=1; //enable external interrupt 0
IT0=1; //select falling edge to trigger external interrupt 0
stop=1; //external interrupt is assigned high potential
b=0; //the parameters used in the interrupt are assigned initial value 0, and each time the interrupt function is entered b++
c=0xfe;
d=1; //the parameter is assigned value 1 to make the main function display run and the digital tube scroll to display 2019


}void keyscan(){
if(key==0)
{
delay(5); //debounceif
(key==0){
stop=0; //P3.2 key is assigned low level when pressed, and the falling edge triggers external interrupt 1
d=0; //display running condition parameters are set to 0, and the display function is terminatedwhile
(!key);
}
}
}void shan() //led light flashing function
{
P1=c; //led light turns on
delay(60);
P1=0xff; //turn off
delay(60);
P1=c; //light up
}void zd() interrupt 0 //interrupt function
{
shan(); //turn on the LED
b++;
c=_crol_(c,1); //c《1switch (b) //according to the value of b (the number of interrupts displayed is different)
{case 1:
dula=1; //turn on segment select
P0=0x06; //assign
dula=0; //turn off segment select
wela=1; //turn on bit select
P0=0xc0; //assign
wela=0; //turn off bit select
break;case 2: dula=1;
P0=0x5b;
dula=0;
wela=1;
P0=0xc0;
wela=0;
break;

case 3: dula=1;
P0=0x4f;
dula=0;
wela=1;
P0=0xc0;
wela=0;
break; case 4: dula=1;
P0=0x66;
dula=0;
wela=1;
P0=0xc0;
wela=0;
break;

case 5: dula=1;
P0=0x6d;
dula=0;
wela=1;
P0=0xc0;
wela=0;
break; case 6: dula=1;
P0=0x7d;
dula=0;
wela=1;
P0=0xc0;
wela=0;
break;

case 7: //显示学号 while(b==7)
{dula=1;
P0=0x5b;
dula=0;
wela=1;
P0=0x3e;
wela=0;
delay(1); dula=1;
P0=0x06;
dula=0;
wela=1;
P0=0x3d;
wela=0;
delay(1); dula=1;
P0=0x07;
dula=0;
wela=1;
P0=0x3b;
wela=0;
delay(1); dula=1;
P0=0x06;
dula=0;
wela=1;
P0=0x37; wela
=0;
delay(1); dula = 1;
P0 = 0x5b ; dula=0; wela = 1;








P0=0xdf;
wela=0;
delay(1);
if(key==0) //Key recognition
{
delay(10); //Debounceif
(key==0){ //Confirm that the key is pressedd
=1; //Reset the display condition parameter to 1, and the display function will runb
=0; //Set b to 0 and recalculate the number of interruptsbreak
;
while(!key);
}
}
}}stop=1; //P3.2 is set to high level}
void main(){
init();
while(1){
keyscan(); //Scan the keyboardif
(d==1) //Display 2019
{
display();
}
keyscan(); //Scan the keyboard
}
}

Problem: After pressing a key, the interrupt cannot be triggered to enter the interrupt function (occasionally, pressing it a few more times can trigger it). After commenting out the display function in the main function, the key triggers the interrupt without any problem. Can someone help me figure out what the problem is?

liwing666

Many of the displays in the middle are a bit cumbersome and seem very long. Can you guys help me look at the problem of interrupt triggering here? All the middle ones are compiled and run without problems. Problem: After pressing the key, the interrupt cannot be triggered to enter the interrupt function (occasionally pressing it a few more times can trigger it). After commenting out the display function in the main function, the key triggering interrupt is no problem. Can you guys help me see what the problem is?

If you have any better ideas for this topic, I hope you can analyze it. Appreciate it! !

damiaa

Your display is too time-consuming. Put it in the interrupt. Keyscan(); takes a long time to execute once. You can't program like this.

liwing666

damiaa posted on 2020-8-8 12:39 Your display is too time-consuming. Put it in the interrupt. keyscan(); takes a long time to execute once. You can't program like this.

I don't quite understand. Isn't keyscan running in a loop in the while statement of the main function? Why does it only run once after a long time?

damiaa

liwing666 posted on 2020-8-29 08:59 I don't quite understand. Isn't keyscan running in a loop in the while statement of the main function? Why does it only run once in a long time?

It means that keyscan() does not need to be run all the time. It should be run for a longer period of time, such as putting it in a timer interrupt. In this way, it can scan once every 10ms or 20ms. No time is wasted.

liwing666

damiaa posted on 2020-8-30 12:48 It means that keyscan() does not need to be run all the time. It should be run for a longer period of time, such as putting it in a timer interrupt. In this way, it can be scanned once every 10ms or 20ms. ...

Senior, shouldn't the more the keyboard is scanned, the more sensitive it is to key presses?

damiaa

liwing666 posted on 2020-9-2 08:06 Senior, shouldn't the more the keyboard is scanned, the more sensitive it is to key presses?

That's not the case.

CPU US-level speed

How sensitive can humans be? Even if they are sensitive, it is only at the 100ms level.