微控制器8051,带矩阵键盘和LED显示屏

时间:2017-03-23 12:43:50

标签: keyboard led keil 8051

我有这样的程序,它有效。我从LED显示屏上的按下按钮获得数字。但是当我按*或#时,我需要更改此程序,以便在显示屏上显示最后2个按下的数字。 例如,我按'1 2 3 4 5#'。在显示器上,我只看到最后两个数字'4 5'。我怎么能这样做?

#include <REGX52.h>
#define SEG P1
#define keypad P2

sbit r1 = P2^0; 
sbit r2 = P2^1; 
sbit r3 = P2^2; 
sbit r4 = P2^3; 

sbit c1 = P2^4; 
sbit c2 = P2^5; 
sbit c3 = P2^6; 
sbit c4 = P3^7;

void scan(void);

unsigned int Display[12] = {0x0, 0x1, 0x2, 0x3,0x4,0x5,0x6,0x7,0x8,0x9};

void main(void)
{
while(1)
{
    scan();
}
}

void scan(void){
r1=0;
r2=r3=r4=1;

if(c1==0)
{
    while(c1==0){
        P1=Display[1];
    }
}
    if(c2==0)
    {
        while(c2==0){
            P1=Display[2];
}
}
    if(c3==0)
{
    while(c3==0){
        P1=Display[3];
    }
}
r2=0;
r1=r3=r4=1;
if(c1==0)
{
    while(c1==0){
        P1=Display[4];
    }
}
if(c2==0)
    {
        while(c2==0){
            P1=Display[5];
}
}
    if(c3==0)
{
    while(c3==0){
        P1=Display[6];
    }
}
r3=0;
r1=r2=r4=1;
if(c1==0)
{
    while(c1==0){
        P1=Display[7];
    }
}
if(c2==0)
    {
        while(c2==0){
            P1=Display[8];
}
}
    if(c3==0)
{
    while(c3==0){
        P1=Display[9];
    }
}
r4=0;
r1=r2=r3=1;
if(c2==0)
{
    while(c2==0){
        P1=Display[0];
}
    }

schema

1 个答案:

答案 0 :(得分:0)

unsigned int last_two_buttons[2] = {0x0, 0x0}; /* array of 2 elements to remember the last two buttons pressed. */

unsigned int update_display = 0; //flag to indicate if LED display needs to be updated.

现在,您不必为每个按下的按钮分配P1 = Display [x],而只需记住/存储按此按钮的按钮,如下所示:

last_two_buttons[0] = last_two_buttons[1]; 
last_two_buttons[1] = Display[x]; //x here indicates the button pressed, the same way as you have been using in your code.

现在,增强scan()以检测*和#按钮。

r4=0;
r1=r2=r3=1;
if(c1==0)
{
    while(c1==0){
        update_display = 1; // * button pressed
}

if(c3==0)
{
    while(c3==0){
        update_display = 1; // # button pressed
}

if(update_display)
{

    P1 = last_two_buttons[0] <<4 + last_two_buttons[1];
    update_display = 0; //reset the variables for next scan.
    last_two_buttons[0] = 0; 
    last_two_buttons[1] = 0;

}

这里的假设是,如果用户只按下一个按钮,比如5#,那么我们将显示0和5作为最后两个按下的按钮。

希望这有帮助。