Question

我希望5 * 5 Led-矩阵中的两个Led（Led 1和Led 3）点亮，并且当按下5 * 5按钮中的相应按钮时，LED的指示灯熄灭，另外两个Led（Led 3和LED 5）应该亮起当按下相应的按钮时，LED指示灯熄灭，另外两个LED指示灯（LED 5和LED 11）应点亮，并且该过程继续进行下一组LED指示灯（LED 11和LED 13），下一组LED指示灯（LED 13和LED 13）。 Led 15）和下一组是（Led 15和Led 17）和（Led 17和Led 21）和（Led 21和Led 23）和（Led 23和Led 25）。

我写的代码是：

/*  If both "first_key_pressed" and "second_key_pressed", we will make the next two LEDs glow and set these two values to zero again
    Initially, LEDs 1 and 3 are glowing, so it is declared that the Led 1 is currently first_in_position and Led 3 is currently second_in_position.
    Once 1 and 3 are pressed, we will change this two values to first_in_position = '3' and second_in_position = '5'
    The method set_next_expected_leds does this job.
*/


#include <Keypad.h>

const byte rows = 5; 
const byte cols = 5; 

char keys[rows][cols] = {
  {'1','2','3','4','5'},
  {'6','7','8','9','A'},
  {'B','C','D','E','F'},
  {'G','H','I','J','K'},
  {'L','M','N','O','P'}
};

byte rowPins[rows] = {40, 39 , 38 , 37 , 36 }; 
byte colPins[cols] = { 35 , 34 ,33 , 32 , 31 }; 

Keypad keypad = Keypad( makeKeymap(keys), rowPins, colPins, rows, cols );

int Led_Rows[] = { 2 , 3 , 4  ,5 ,6 } ;
int Led_Columns[] = { 7 , 8 , 9 , 10 , 11 };
int Loop_Count = 5 ;
int first_key_pressed = 0;
int second_key_pressed = 0;
char first_in_position = '1';
char second_in_position = '3';
int Wait_Time = 1000 ; 
int i = 0 ;
int j = 0 ;
int x = 0 ;
int y = 0 ; 
int c = 0 ;

void setup() {
  Serial.begin(9600);

  for(int i=0;i<5;i++){
    pinMode(Led_Rows[i], OUTPUT);
    digitalWrite(Led_Rows[i],LOW);
    pinMode(Led_Columns[i], OUTPUT);
    digitalWrite(Led_Columns[i],HIGH);
  }
  Led_Glow(0,0);
  Led_Glow(0,2);
}


void All_On(){
 for( i = 0  ; i < Loop_Count  ; i++ ){
    digitalWrite(Led_Rows[i],HIGH);
    while( j < Loop_Count ){
    digitalWrite(Led_Columns[j],LOW);
    j++ ;
    }
  }
  j = 0 ;
}

void All_Off(){
  for(  i = 0 ; i < Loop_Count ; i++ ){
    digitalWrite(Led_Rows[i],LOW);
    digitalWrite(Led_Columns[i],HIGH);
  }
}

void Led_Glow( int a , int b ){
  digitalWrite(Led_Rows[a],HIGH);
  digitalWrite(Led_Columns[b],LOW); 
}

void Led_Dim( int a ,int b ){
  digitalWrite(Led_Rows[a],LOW);
  digitalWrite(Led_Columns[b],HIGH);
}


void check_for_position(int c) {
  if (first_in_position == c) {
    first_key_pressed = 1;
  } else if (second_in_position == c) {
    second_key_pressed = 1;
  }
}

void set_next_expected_leds() {
  if (first_in_position == '1') {
    first_in_position = '3';
    second_in_position = '5';
    Led_Glow(0,2);
    Led_Glow(0,4);
  } else if (first_in_position == '3') {
    first_in_position = '5';
    second_in_position = 'B';
    Led_Glow(0,4);
    Led_Glow(2,0);
  } else if (first_in_position == '5') {
    first_in_position = 'B';
    second_in_position = 'D';
    Led_Glow(2,0);
    Led_Glow(2,2);
  } else if (first_in_position == 'B') {
    first_in_position = 'D';
    second_in_position = 'F';
    Led_Glow(2,2);
    Led_Glow(2,4);
  } else if (first_in_position == 'D') {
    first_in_position = 'F';
    second_in_position = 'L';
    Led_Glow(2,4);
    Led_Glow(4,0);
  } else if (first_in_position == 'F') {
    first_in_position = 'L';
    second_in_position = 'N';
    Led_Glow(4,0);
    Led_Glow(4,2);
  } else if (first_in_position == 'L') {
    first_in_position = 'N';
    second_in_position = 'P';
    Led_Glow(4,2);
    Led_Glow(4,4);
  } else if (first_in_position == 'N') {

  } else if (first_in_position == 'P') {

  }

  first_key_pressed = 0;
  second_key_pressed = 0;
}

void loop() { 
  char key = keypad.getKey();
   switch (key) {
    case '1':
      check_for_position('1');
      Led_Dim(0, 0);
      break;

    case '3':
      check_for_position('3');
      Led_Dim(0, 2);
      break;

    case '5':
      check_for_position('5');
      Led_Dim(0, 4);
      break;

    case 'B':
      check_for_position('B');
      Led_Dim(2, 0);
      break;

    case 'D':
      check_for_position('D');
      Led_Dim(2, 2);
      break;

    case 'F':
      check_for_position('F');
      Led_Dim(2, 4);
      break;

    case 'L':
      check_for_position('L');
      Led_Dim(4, 0);
      break;

    case 'N':
      check_for_position('N');
      Led_Dim(4, 2);

      break;

    case 'P':
      check_for_position('P');
      Led_Dim(4, 4);
    }
   if(first_key_pressed == 1 && second_key_pressed == 1) {
    set_next_expected_leds();
  }
}

预期结果：如果用户按下按钮1，则可以在第一步中按下按钮1或按钮3，LED-1应该熄灭，而在按下按钮3之后，LED-3应该熄灭，并且另一组LED（LED -3和Led-5）应该发光。

注意：用户有权先按下按钮3，然后再按下按钮1，然后再先按下led-3，然后再按下led-1，然后下一组LED才能发光。

实际结果：如果我按下按钮1，则led-1和led-3都会熄灭，并且当我按下Button-3时，一些随机的led也会开始闪烁。

如何通过5 * 5按钮控制5 * 5 LED-矩阵中的LED

0 个答案: