Problem saving data on an array

Projects Programming
1

Hello, I'm currently working on this project. The objective it's introduce values with a keypad and save it in the array 'r[3][6]', then when i press the key 'S' it will compare the values of the internal clock with the array values, it works like a chronometer.
Using the code without comparing values array-time, the serial port shows me that this part is working okay. Like the first picture.
But when it try to compare the values that the user introduced with the real time of the internal clock doesn't work, it's shown in the second picture. For me the logic it's okay, but i don't if this could be possible.

Here is the code:

#include <Time.h>
#include <TimeLib.h>
#include <Keypad.h>

const byte rows = 4;
const byte cols = 3;

char keys[rows][cols] = {
{'1', '2', '3'},
{'4', '0', 'S'},
{'7', '8', '9'},
{'E', '0', 'S'}
};
byte row[rows] = {13,12,11,10};
byte col[cols] = {9,8,7};
Keypad keypad =Keypad(makeKeymap(keys), row, col, rows, cols);

int r[3][6];
int i=0, a=0, c=0, u=0;
char j, z;
int h=0;
int m=00;
int s=00;
int n, hd, hu, H, md, mu, M, sd, su, S;

void setup() {
Serial.begin(9600);
setTime(h, m, s, 0, 0, 0);//hora, minutos, segundos, dia, mes,año
pinMode(3,OUTPUT);
}

void loop() {

char key =keypad.getKey();
time_t t = now();

//Adding values

if(key!=NO_KEY){
n=num(key);
Serial.print(n);
r*[a] = n;*

  • i++;*

  • a++;*

  • j++;*

  • if (j==2)*

  • {*

  • Serial.print(":");*

  • }*

  • if(j==4){*

  • Serial.print(":");*

  • }*

  • if(j==6){*

  • Serial.println();*

  • j=0;*

  • }*

  • delay(500);*

  • }*

  • if(key =='S'){*

  • delay(10);*

  • u++;*

  • }*

  • //Comparing values*

  • if(u==1){*

  • //Horas*

  • hd =((r*
    ```c
    *[0])*10);
        hu =((r[c][1])*1);
        H=hd+hu;
        //Minutos
        md =((r[c][2])*10);
        mu =((r[c][3])*1);
        M=md+mu;
        //Segundos
        sd =((r[c][4])*10);
        su =((r[c][5])*1);
        S=sd+su;
        if(H==hour(t) && M==minute(t) && S==second(t)){
          digitalWrite(3,HIGH);
          delay(1000);
          digitalWrite(3,LOW);
          delay(1000);
          digitalWrite(3,HIGH);
          delay(1000);
          digitalWrite(3,LOW);
          delay(1000);
          c++;
        }
        if(c==3){
          c=0;
          u=0;
        }
      }
    }

char num(char z){
  int result;
  switch(z){
    case '1':
    result=1;
    break;
    case '2':
    result=2;
    break;
    case '3':
    result=3;
    break;
    case '4':
    result=4;
    break;
    case '5':
    result=5;
    break;
    case '6':
    result=6;
    break;
    case '7':
    result=7;
    break;
    case '8':
    result=8;
    break;
    case '9':
    result=9;
    break;
    case '0':
    result=0;
    break;
  }
  return result;
}

Anotación 2020-05-31 175110.png*
```

2 
char num(char z){
  int result;
  switch(z){
    case '1':
    result=1;
    break;
    case '2':
    result=2;
    break;
    case '3':
    result=3;
    break;
    case '4':
    result=4;
    break;
    case '5':
    result=5;
    break;
    case '6':
    result=6;
    break;
    case '7':
    result=7;
    break;
    case '8':
    result=8;
    break;
    case '9':
    result=9;
    break;
    case '0':
    result=0;
    break;
  }
  return result;
}

In other words:

char num(char z){
  if (isDigit(z))
    return z - '0';
  else
    return 0;
}
3

This is an odd keypad - no '5' or '6'?

char keys[rows][cols] = {
  {'1', '2', '3'},
  {'4', '0', 'S'},
  {'7', '8', '9'},
  {'E', '0', 'S'}
};

You don't need a two dimensional array to store your input values. Your code never resets the 'i' or 'j' variables so after 3 digits, you are writing out of bounds.