51單片機的智能密碼鎖_51單片機的電子密碼鎖程序用KEIL C寫的

㈠如何利用51單片機設計一個電子密碼鎖系統

用單片機最小系統，外擴一個4*4矩陣鍵盤，再加一個顯示設備，蜂鳴器，就可以設計出一個密碼鎖系統了。

㈡ 51單片機的電子密碼鎖程序，用KEIL C寫的

以前幫別人做過一個類似的

#include<reg52.h>

#include <def.h>

#include"lcd.c"

uchar key;

uchar num;

sbit alarm=P2^4;

sbit open=P2^1;

uchar code table[]={' ','1','2','3',

'4','5','6','7',

'8','9','0'};

void keyscan();

uchar hptable[10];

uchar password[6];

uchar wordbuf[6];

uchar fom[]="input password:";

uchar change[]="set mask:";

uchar state1[]="right";

uchar state2[]="error";

uchar sflag;

uchar dflag,cnt;

uchar setf,fset;

uchar flag;

void main ()

{

uchar hp;

alarm=0;

flag=0;

setf=0;

fset=0;

dflag=0;

lcden=0;

lcdrw=1;

lcdrs=1;

P3=0x0f;

for(hp=0;hp<6;hp++)

{

password[hp]='8';

}

write(0x38,0); //0為寫命令

delay(9);

write(0x08,0);

write(0x01,0);

write(0x06,0);

write(0x0c,0);

write(0x80,0);

for(hp=0;hp<15;hp++)

{

write(fom[hp],1);

}

while(1)

{

// alarm=!alarm;

keyscan();

if(setf==0)

{

if(num>=1&&num<=10)

{

write(0x80+0x40+cnt,0);

write('*',1);

wordbuf[cnt]=table[num];

cnt++;

num=0;

}

else

{

write(0x01,0);

write(0x80,0);

for(hp=0;hp<9;hp++)

{

write(change[hp],1);

}

fset=1;

cnt=0;

while(fset)

{

keyscan();

if(num>=1&&num<=10)

{

write(0x80+0x40+cnt,0);

write('*',1);

password[cnt]=table[num];

cnt++;

num=0;

}

if(num==11)

{

write(0x80+0x40+cnt-1,0);

write('',1);

cnt--;

num=0;

}

if(num==13)

{

flag=1;

num=0;

fset=0;

setf=0;

// goto chuz;

}

cnt=0;

}

if(flag==1)

{

write(0x01,0);

write(0x80,0);

for(hp=0;hp<15;hp++)

{

write(fom[hp],1);

}

flag=0;

}

if(num==11)

{

write(0x80+0x40+cnt-1,0);

write('',1);

cnt--;

num=0;

}

if(num==12)

{

if(dflag==1)

setf=1;

num=0;

}

if(num==13)

{

num=0;

write(0x80+0x40+10,0);

if(wordbuf[0]==password[0]&&wordbuf[1]==password[1]&&wordbuf[2]==password[2]&&wordbuf[3]==password[3]&&wordbuf[4]==password[4]&&wordbuf[5]==password[5])

{

for(hp=0;hp<5;hp++)

{

write(state1[hp],1);

}

sflag=0;

dflag=1;

}

else

{

for(hp=0;hp<5;hp++)

{

write(state2[hp],1);

}

sflag++;

}

if(sflag==3)

{

alarm=!alarm;

//delay(9);

}

if(num==14)

{

num=0;

cnt=0;

write(0x01,0);

write(0x80,0);

for(hp=0;hp<15;hp++)

{

write(fom[hp],1);

}

if(num==15)

{

if(dflag==1)

{

open=0;

dflag=0;

}

void keyscan ()

{

uchar tem,temp;

P3=0xfe;

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0) //沒有鍵按下時它的值是不變的，即仍為0XF0，而有鍵接下時要變

{

delay(5); //延時消抖

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0) //確認有鍵按下

{

temp=P3;

tem=temp&0xf0; //這里也可以不要這一步，但下面的case里的值也要變

switch (tem)

{

case 0xe0: num=0; break;

case 0xd0: num=1; break;

case 0xb0: num=2; break;

case 0x70: num=3; break;

default : ; break ;

}

while(tem!=0xf0) //等待鬆手

{

temp=P3;

tem=temp&0xf0;

}

P3=0xfd;

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0)

{

delay(5);

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0)

{

temp=P3;

tem=temp&0xf0;

switch (tem)

{

case 0xe0: num=4; break;

case 0xd0: num=5; break;

case 0xb0: num=6; break;

case 0x70: num=7; break;

default : ; break ;

}

while(tem!=0xf0)

{

temp=P3;

tem=temp&0xf0;

}

P3=0xfb;

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0)

{

delay(5);

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0)

{

temp=P3;

tem=temp&0xf0;

switch (tem)

{

case 0xe0: num=8; break;

case 0xd0: num=9; break;

case 0xb0: num=10; break;

case 0x70: num=11; break;

default : ; break ;

}

while(tem!=0xf0)

{

temp=P3;

tem=temp&0xf0;

}

P3=0xf7;

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0)

{

delay(5);

temp=P3;

tem=temp&0xf0;

if (tem!=0xf0)

{

temp=P3;

tem=temp&0xf0;

switch (tem)

{

case 0xe0: num=12; break;

case 0xd0: num=13; break;

case 0xb0: num=14; break;

case 0x70: num=15; break;

default : ; break ;

}

while(tem!=0xf0)

{

temp=P3;

tem=temp&0xf0;

}

㈢ 51單片機密碼鎖的原理是什麼

簡單密碼鎖，掃描按鍵，獲得6位或8位數字，與事先設定好的密碼比較，相同則開鎖，不同則返回或報警。簡單的程序密碼可以直接寫進程序中。

㈣基於51單片機的密碼鎖

基於51單片機的密碼鎖的話其實我是不知道是什麼意思的所以說不好意思咯(๑ó﹏ò๑)

㈤ 51單片機關於密碼鎖的畢業設計，論文

程序設計內容

（1）． 密碼的設定，在此程序中密碼是固定在程序存儲器ROM中，假設預設的密碼為「12345」共5位密碼。

（2）． 密碼的輸入問題：由於採用兩個按鍵來完成密碼的輸入，那麼其中一個按鍵為功能鍵，另一個按鍵為數字鍵。在輸入過程中，首先輸入密碼的長度，接著根據密碼的長度輸入密碼的位數，直到所有長度的密碼都已經輸入完畢；或者輸入確認功能鍵之後，才能完成密碼的輸入過程。進入密碼的判斷比較處理狀態並給出相應的處理過程。

（3）．按鍵禁止功能：初始化時，是允許按鍵輸入密碼，當有按鍵按下並開始進入按鍵識別狀態時，按鍵禁止功能被激活，但啟動的狀態在3次密碼輸入不正確的情況下發生的。

C語言源程序

#includeunsignedcharcodeps[]={1,2,3,4,5};

unsignedcharcodedispcode[]={0x3f,0x06,0x5b,0x4f,0x66,                              

0x6d,0x7d,0x07,0x7f,0x6f,0x00,0x40};

unsignedcharpslen=9;unsignedchartemplen;

unsignedchardigit;unsignedcharfuncount;

unsignedchardigitcount;

unsignedcharpsbuf[9];

bitcmpflag;

bithibitflag;

biterrorflag;

bitrightflag;

unsignedintsecond3;

unsignedintaa;

unsignedintbb;

bitalarmflag;

bitexchangeflag;

unsignedintcc;

unsignedintdd;

bitokflag;

unsignedcharoka;

unsignedcharokb;

voidmain(void)

{ 

unsignedchari,j; 

P2=dispcode[digitcount]; 

TMOD=0x01; 

TH0=(65536-500)/256; 

TL0=(65536-500)%6; 

TR0=1; 

ET0=1; 

EA=1; 

while(1)  

 {     

if(cmpflag==0)       

{         

if(P3_6==0)//functionkey          

 {             

for(i=10;i>0;i--)             

for(j=248;j>0;j--);     

        if(P3_6==0)               

{               

  if(hibitflag==0)      

             {    

                 funcount++; 

                    if(funcount==pslen+2)

                       { 

                        funcount=0;

                         cmpflag=1;

                        }

                      P1=dispcode[funcount];

                   }

                   else

                     {

                        second3=0;

                     } 

                while(P3_6==0);

               }

         if(P3_7==0)//digitkey

           {

             for(i=10;i>0;i--)

             for(j=248;j>0;j--);

             if(P3_7==0)

               {

                 if(hibitflag==0)

                   {

                     digitcount++; 

                 if(digitcount==10)

                       {

                         digitcount=0;

                       }

                     P2=dispcode[digitcount];

                     if(funcount==1)

                       {

                         pslen=digitcount;                         

templen=pslen;

                       }

                       elseif(funcount>1)

                         { 

                          psbuf[funcount-2]=digitcount;

                         }

                   else

                     {

                       second3=0;

                     }

                 while(P3_7==0);

               }

      else

         {

           cmpflag=0;

           for(i=0;i

             { 

              if(ps[i]!=psbuf[i])

                 {

                   hibitflag=1;

                   i=pslen;

                   errorflag=1;

                   rightflag=0;

                   cmpflag=0;

                   second3=0;

                   gotoa; 

                }

         cc=0; 

          errorflag=0; 

          rightflag=1;

           hibitflag=0;

a:   cmpflag=0;

         }

voidt0(void)

interrupt1using0{ TH0=(65536-500)/256; 

TL0=(65536-500)%6; 

if((errorflag==1)&&(rightflag==0)) 

{

     bb++;

     if(bb==800)

       {

         bb=0;

         alarmflag=~alarmflag;

       }

     if(alarmflag==1)

       {

         P0_0=~P0_0;

       }

     aa++;

     if(aa==800)

       {

         aa=0;

         P0_1=~P0_1;

       }

     second3++;

     if(second3==6400)

       {

         second3=0;

         hibitflag=0;

         errorflag=0;

         rightflag=0;

         cmpflag=0;

         P0_1=1; 

        alarmflag=0;

         bb=0; 

        aa=0; 

      }

 if((errorflag==0)&&(rightflag==1))

   {

     P0_1=0;

     cc++;

     if(cc<1000)

       {

         okflag=1;

       }

       elseif(cc<2000)

         {

           okflag=0;

         }

         else

           {

             errorflag=0;

             rightflag=0;

             hibitflag=0;

             cmpflag=0;

             P0_1=1;

             cc=0; 

            oka=0;

             okb=0;

             okflag=0; 

            P0_0=1; 

          }

     if(okflag==1)

       { 

        oka++; 

        if(oka==2)

           {

             oka=0;

             P0_0=~P0_0;

           }

       else

         {

           okb++;

           if(okb==3)

             {

               okb=0;

               P0_0=~P0_0;

             } 

㈥基於51單片機的指紋密碼鎖lcd1602顯示

有七51單片機的指紋密碼鎖，這個屏不認識，1602顯示你這可以上網咨詢一下，到。怎麼回就知

㈦怎麼設計一個由51單片機控制的電子密碼鎖

#include<reg51.h>

#defineuintunsignedint

charleab[]={0x7f,0x7f,0x7f,0x7f,0x7f,0x7f};

delay(uintz)

{

uintx;

for(;z<0;z--)

for(x=0;x<10;x++);//1ms

}

disp()

{

charx,y;

for(x=5;x>=0;x--)

{for(y=0;y<10;y++)

{P0=0;

P2=x;

P0=leab[x];

delay(10);

P0=0;}

}

charsao(charg)

{chard;

switch(g)

{

case1:d=1;break;

case2:d=2;break;

case4:d=3;break;

case8:d=4;break;

default:d=0;break;

}

returnd;

}

charkey()

{

charH=0,L=0,d,d1,d2;

while(1)

{d=0;

P1=0xf0;

if(P1!=0xf0)

{

delay(10);

if(P1!=0xf0)

{

H=(~(P1^0x0f))>>4;

P1=0xff;

P1=0x0f;

L=~(P1^0xf0);

d2=sao(L);

d1=sao(H);

d=(d1-1)*4+d2;

P2=d;

}}

if(d!=0)break;

disp();

}

returnd;

}

main()

{

charleab1[]={0x00,0x00,0x00,0x00,0x00,0x00};

charleab2[]={0x09,0x08,0x00,0x06,0x01,0x01};

chara,b,d,e=0;

intf;

while(1)

{

P3=0xff;

for(a=0;a<6;a++)

{

leab[a]=0x7f;

leab1[a]=0x00;

}

for(b=0;b<100;b++)

{

disp();

}

//for(c=0;c<6;c++)

//{

//leab[c]=0x00;

//}

while(1)

{

for(d=0;d<6;d++)

{

e=key();

if((e>=1)&(e<=10))

{

leab[d]=0x71;

disp();

leab1[d]=e-1;

}

else

{d=d-1;disp();}

if(e==15)break;

}

if(e==15)break;

while(e!=16)

{disp();

e=key();

if(e==16)break;

if(e==15)break;

}

if(e==15)break;

if((leab1[0]==leab2[0])&(leab1[0]==leab2[0])&

(leab1[0]==leab2[0])&(leab1[0]==leab2[0])&

(leab1[0]==leab2[0])&(leab1[0]==leab2[0]))

{

for(f=0;f<800;f++)

{P3=0x02;

delay(100);

disp();

}

else

{for(f=0;f<800;f++)

{P3=0x01;

delay(100);

disp();

}}

break;

}

}}

㈧求一個51單片機的簡易密碼鎖程序

你好！

從你的功能要求來看，這個密碼鎖不簡易，功能很強很全了；

你這個是模擬驗證程序還是要做出實物來驗證程序。

採用模擬軟體模擬出的密碼鎖

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