這個代碼是0~99 基本都是按一下+1或者-1我覺得好像只有用定時器0和1才可以弄這樣子，但是，代碼是怎么寫？謝謝大家了。我因為我腦子問題太多，總在想，按1下有點慢。換長按，這樣連續快速，才可以連續加1或者減1了。
單片機源程序如下:
- #include "reg52.h"
- typedef unsigned int u16;
- typedef unsigned char u8;
- sbit LSA=P2^2;
- sbit LSB=P2^3;
- sbit LSC=P2^4;
- sbit k1=P3^1;
- sbit k2=P3^0;
- sbit k3=P3^2;
- sbit k4=P3^3;
- u8 code smgduan[17]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,
- 0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};//顯示0~F的值
- #define KEY1 1
- #define KEY2 2
- #define KEY3 3
- #define KEY4 4
- char keynum;
- u8 DisplayData[8];void delay(u16 i)
- {
- while(i--);
- }
- u8 KeyScan(u8 mode)
- {
- static u8 keyen=1;
- if(mode==1)
- {
- keyen=1;
- }
- if(keyen==1&&(k1==0||k2==0||k3==0||k4==0))
- {
- delay(1000); //消抖處理
- keyen=0;
- if(k1==0)return KEY1;
- else if(k2==0)return KEY2;
- else if(k3==0)return KEY3;
- else if(k4==0)return KEY4;
- }
- else if(k1==1&&k2==1&&k3==1&&k4==1)
- {
- keyen=1;
- }
- return 0;
- }void DigDisplay()
- {
- u8 i;
- for(i=0;i<8;i++)
- {
- switch(i)
- {
- case(0):
- LSA=0;LSB=0;LSC=0; break
- case(1):
- LSA=1;LSB=0;LSC=0; break
- case(2):
- LSA=0;LSB=1;LSC=0; break
- case(3):
- LSA=1;LSB=1;LSC=0; break
- case(4):
- LSA=0;LSB=0;LSC=1; break
- case(5):
- LSA=1;LSB=0;LSC=1; break
- case(6):
- LSA=0;LSB=1;LSC=1; break
- case(7):
- LSA=1;LSB=1;LSC=1; break
- }
- P0=DisplayData[i];
- delay(100);
- P0=0x00;
- }
- }
- void datapros()
- {
- u8 key=KeyScan(0);
- switch(key)
- {
- case 1: keynum++;if(keynum==100)keynum=0; break;
- case 2: keynum--;if(keynum<=0)keynum=99; break;
- case 3: keynum=0; break;
- case 4: break;
- }
- DisplayData[0]=smgduan[keynum%10];
- DisplayData[1]=smgduan[keynum/10];void main()
- {
- while(1)
- {
- datapros();
- DigDisplay();
- }
- }
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