标题: 单片机计算器程序 [打印本页]

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作者: 金华佳哥哥    时间: 2021-8-25 15:22
标题: 单片机计算器程序
简易计算器有图加程序

1. #include"key16a.h"
   
2. #include"lcd1602.h"
   
3. Uchar code keyval[16]={'/','=',0,'C','*',9,8,7,'-',6,5,4,'+',3,2,1,};
   
4. Uchar data numb1[4],numb2[4],result[12];
   
5. void Fclear ()
   
6.         {Uchar i;
   
7. for(i=0;i<4;i++)
   
8. { numb1[i]=0;
   
9.         numb2[i]=0;
   
10. }
    
11. for(i=0;i<12;i++)
    
12. result[i]=' ';
    
13. }
    
14. void main()
    
15. { Uchar i=0,j=0,tmp=0;
    
16. Uchar oprat1=0,oprat2=0,opratflag=0;
    
17.         Uint answer=0;
    
18.         Fclear ();
    
19. RstLcd();
    
20.         ClrLcd();
    
21.         while(1)
    
22.         { do{tmp=key_scan();}while(tmp==0xff);
    
23. tmp=keyval[tmp];
    
24. if(tmp=='C'){ClrLcd();break;}
    
25. else
    
26. {if(tmp!='=')
    
27. { if((tmp<10)&&(opratflag==0))
    
28. { numb1[i]=tmp;WriteChar(tmp+0x30,i+4,0);i++;}
    
29. else if (tmp>10)
    
30. {opratflag=tmp;WriteChar(tmp,7,0);}
    
31. else
    
32. {numb2[j]=tmp;WriteChar(tmp+0x30,j+8,0);j++;}
    
33. }
    
34. else
    
35. {
    
36. switch(i)
    
37. { case 1: oprat1=numb1[0];
    
38. numb1[0]+=0x30;numb1[1]=numb1[2]=' ';
    
39.         break;
    
40.         case 2:
    
41. oprat1=numb1[0]*10+numb1[1];
    
42.         numb1[0]+=0x30;numb1[1]+=0x30;
    
43.         numb1[2]=' ';break;
    
44.         case 3:
    
45. oprat1=numb1[0]*100+numb1[1]*10+numb1[2];
    
46. numb1[0]+=0x30;numb1[1]+=0x30;
    
47.         numb1 [2]+=0x30;
    
48.         break;
    
49. }
    
50.         switch(j)
    
51. { case 1: oprat2=numb2[0];
    
52. numb2[0]+=0x30;numb2[1]=numb2[2]=' ';
    
53.         break;
    
54.         case 2:
    
55. oprat2=numb2[0]*10+numb2[1];
    
56.         numb2[0]+=0x30;numb2[1]+=0x30;
    
57.         numb2 [2]=' '; break;
    
58.         case 3:
    
59. oprat2=numb2[0]*100+numb2[1]*10+numb2[2];
    
60.         numb2[0]+=0x30;numb2[1]+=0x30;
    
61.         numb2 [2]+=0x30;
    
62.         break;
    
63. }
    
64. result [0]=' ';
    
65. switch(opratflag)
    
66. { case '+':
    
67. answer=oprat1+oprat2;break;
    
68. case '-':
    
69. if(oprat1<oprat2)
    
70. { answer=oprat2-oprat1;
    
71.         result[0]='-';}
    
72. else
    
73. {answer=oprat1-oprat2;
    
74. }
    
75. break;
    
76. case'*':
    
77. answer=oprat1*oprat2;break;
    
78. case '/':
    
79. answer=oprat1/oprat2;
    
80. tmp=oprat1%oprat2;
    
81. if(tmp>=100) i=3;
    
82. else if(tmp>9){i=2;}
    
83. else if (tmp>0)
    
84. {i=1;}
    
85. else break;
    
86. result[6]= result[7]=result[8]='.';
    
87. j=9;
    
88. switch(i)
    
89. {case 3:result[j]=tmp/100+0x30;
    
90.         case 2:result[j++]=(tmp/10)%10+0x30;
    
91.         case 1:result[j]=tmp%10+0x30;
    
92. break;
    
93. }
    
94. break;
    
95. }
    
96. if(answer>=10000) i=5;
    
97. else if (answer>=1000)i=4;
    
98. else if (answer>=100)i=3;
    
99. else if(answer>=10)i=2;
    
100. else i=1;
     
101. j=1;
     
102. switch(i)
     
103. {case 5:result[j++]=answer/10000+0x30;
     
104.         case 4:result[j++]=(answer/1000)%10+0x30;
     
105.         case 3:result[j++]=(answer/100)%10+0x30;
     
106.         case 2:result[j++]=(answer/10)%10+0x30;
     
107.         case 1:result[j]=answer%10+0x30;break;
     
108. }
     
109. WriteChar('=', 3,1);
     
110. WriteString(result,4,1);
     
111. i=0;j=0;tmp=0; answer=0;
     
112. oprat1=0,oprat2=0,opratflag=0;
     
113. Fclear();
     
114. }
     
115. }
     
116. }
     
117. }
     

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