標(biāo)題: 單片機(jī)矩陣鍵盤(pán)顯示電路 數(shù)碼管顯示是亂碼的情況,怎么辦 [打印本頁(yè)]
作者: 14150156    時(shí)間: 2023-12-13 22:44
標(biāo)題: 單片機(jī)矩陣鍵盤(pán)顯示電路 數(shù)碼管顯示是亂碼的情況,怎么辦
目的：矩陣鍵盤(pán)顯示電路，采用4*4鍵盤(pán)，用4位數(shù)碼管顯示輸入的數(shù)字，依次將輸入數(shù)字組合在一起顯示

單片機(jī)源程序如下:
  1. #include <reg51.h>
  2. unsigned char  WEI[]={0x01,0x02,0x04,0x08,0x10,0x20};
  3. unsigned char table[]={0xC0,0xF9,0xA4,0xB0,0x99,0x92,0x82,0xF8,0x80,0x90,0x88,0x83,0xc6,0xa1,0x86,0x8e};
  4. unsigned char a[4] = {0,0,0,0};
  5. unsigned char temp;
  6. unsigned char key();
  7. unsigned char i,n,shu;
  8. void delayms(unsigned int ms);
  9. void keyscan(void);
  10. void shuma(unsigned char par,unsigned char dat)
  11. {
  12. P2 = WEI[par];
  13. P0 = table[dat];
  14. }

  16. void main()
  17. {
  18. i = n = 0;
  19.   while(1)
  20.   {
  21.    shu = key();
  22.    if(shu !=17)
  23.    {
  24.     a[n] = shu;
  25.     n++;
  26.     if(n>3) n = 0;
  27.    }
  28.    shuma(i,a[i]);
  29.    i++;
  30.    if(i>3) i = 0;
  31.    delayms(10);
  32.   }
  33. }
  34. unsigned char key()
  35. {
  36. unsigned char key1;
  37. key1 = 17;
  38. P3 = 0xEf;
  39. if (P3 != 0xEf)
  40. {
  41.   delayms(10);
  42.   switch(P3)
  43.   {
  44.    case 0xE7:key1 = 0;break;
  45.    case 0xEB:key1 = 1;break;
  46.    case 0xED:key1 = 2;break;
  47.    case 0xEE:key1 = 3;break;
  48.   }
  49.   while(P3 != 0xEf);
  50. }

  52. P3 = 0xDF;
  53. if (P3 != 0xDF)
  54. {
  55.   delayms(10);
  56.   switch(P3)
  57.   {
  58.    case 0xD7:key1 = 4;break;
  59.    case 0xDB:key1 = 5;break;
  60.    case 0xDD:key1 = 6;break;
  61.    case 0xDE:key1 = 7;break;
  62.   }
  63.   while(P3 != 0xDF);

  65. }

  67. P3 = 0xBF;
  68. if (P3 != 0xBF)
  69. {
  70.   delayms(10);
  71.   switch(P3)
  72.   {
  73.    case 0xb7:key1 = 8;break;
  74.    case 0xBB:key1 = 9;break;
  75.    case 0xBD:key1 = 10;break;
  76.    case 0xBE:key1 = 11;break;
  77.   }
  78.     while(P3 != 0xBF);
  79. }


  82. P3 = 0x7f;
  83. if (P3 != 0x7f)
  84. {
  85. delayms(10);
  86.   switch(P3)
  87.   {
  88.    case 0x77:key1 = 12;break;
  89.    case 0x7B:key1 = 13;break;
  90.    case 0x7D:key1 = 14;break;
  91.    case 0x7E:key1 = 15;break;
  92.   }
  93.   while(P3 != 0x7f);
  94. }

  96. return key1;
  97. }



  101. void delayms(unsigned int ms)
  102. {
  103.   unsigned int x,y;
  104. for (x=ms;x>0;x--)        
  105.   for(y=120;y>0;y--);
  106. }
復(fù)制代碼

作者: Hephaestus    時(shí)間: 2023-12-13 23:37
根本就沒(méi)亂，那么顯眼的“6512”四個(gè)數(shù)字看不見(jiàn)？？？
作者: hhhxxjx    時(shí)間: 2023-12-13 23:52
送你模塊化的程序

3.矩陣按鍵行列掃描法.rar

291.97 KB, 下載次數(shù): 6
作者: hange_v    時(shí)間: 2023-12-14 08:43
Hephaestus 發(fā)表于 2023-12-13 23:37
根本就沒(méi)亂，那么顯眼的“6512”四個(gè)數(shù)字看不見(jiàn)？？？

哈哈    好眼力
作者: lkc8210    時(shí)間: 2023-12-14 09:03
  1. void shuma(unsigned char par,unsigned char dat)
  2. {
  3. P0 = 0xFF;//消影
  4. P2 = WEI[par];
  5. P0 = table[dat];
  6. }
復(fù)制代碼

先試試加消影
感覺(jué)共陰共陽(yáng)弄錯(cuò)了



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