keypad storing problem any idea store more than 11 numbers like phone number

Question

#include <SoftwareSerial.h>
#include <LiquidCrystal.h>

SoftwareSerial mySerial(6,5);
int table[]={0,0,0};

#include <Keypad.h>
#include <Wire.h>

LiquidCrystal lcd(7, 8, 9, 10, 11, 12);
const byte ROWS = 4; //four rows
const byte COLS = 3; //three columns

char keys[ROWS][COLS] = {
    {'1','2','3'},
    {'4','5','6'},
    {'7','8','9'},
    {'*','0','#'}
};
byte rowPins[ROWS] = {2, 3, 17, 16}; //connect to the row pinouts of the keypad
byte colPins[COLS] = {15, 14, 13}; //connect to the column pinouts of the keypad
int count=0;
Keypad kpd = Keypad(makeKeymap(keys), rowPins, colPins, ROWS, COLS);
char entryStr[13];   // This can hold up to 4 digits
int i=0;
***char n1, n2 , n3;
char buf [15] ;// i put here***

void setup()
{
    Wire.begin();
    mySerial.begin(9600);
    Serial.begin(9600);
    lcd.begin(20, 4);
    ask();
}

void loop()
{
    Wire.requestFrom(2, 3);    // request 6 bytes from slave device #2
    for (int i = 0; i < 3; i++)
    {
        int c = Wire.read(); // receive a byte as character
        Serial.print(c);
        table[i] = c;
        Serial.print('\t');
        // print the character
    }
    Serial.print('\n');
    Serial.print(table[0]);
    Serial.print('\t');
    Serial.print(table[1]);
    Serial.print('\t');
    Serial.print(table[2]);
    Serial.print('\n');

    //if(table[0]<=8)
    //{
    //Serial.print("Flame Detected!");
    //Serial.println("Level1");
}

void ask()
{
    lcd.clear();
    lcd.setCursor(0, 0);
    lcd.print("Enter no1.:");
    lcd.setCursor(0,1);
    lcd.print("(starts with 0)");
    lcd.setCursor(1, 2);
    n1 = GetNumber(buf, sizeof (buf) - 1);

    Serial.print(n1);

    delay(1);

    lcd.clear();
    lcd.setCursor(0, 0);
    lcd.print("Enter no2.:");
    lcd.setCursor(0, 1);
    lcd.print("(starts with 0)");
    lcd.setCursor(1, 2);
    n2 = GetNumber(buf, sizeof (buf) - 1);

    Serial.println(n2);
    delay(1);
    lcd.clear();
    lcd.setCursor(0,0);
    lcd.print("Enter no3.:");
    lcd.setCursor(0,1);
    lcd.print("(starts with 0)");
    lcd.setCursor(1,2);
    n3 = GetNumber(buf, sizeof (buf) - 1);

    Serial.println (n3);
    delay(1);
    Serial.println (n1);
    Serial.println (n2);
    Serial.println (n3);
    lcd.clear();
    lcd.setCursor(0, 0);
    lcd.print("list of numbers ");
    lcd.setCursor(0, 1);
    lcd.print("will be shown press");
    lcd.setCursor(1, 2);
    lcd.print("reset if wrong");


    delay(7000);
    lcd.clear();
    lcd.setCursor(0, 0);
    lcd.print("Numbers Stored:");
    lcd.setCursor(0, 1);
    lcd.print("no1.:0");
    lcd.print(n1);
    lcd.setCursor(1, 2);
    lcd.print("no2.:0");
    lcd.print(n2);
    lcd.setCursor(2, 3);
    lcd.print("no3.:0");
    lcd.print(n3);
}

void GetNumber(char * result, const int maxLength)
{
    char key = kpd.getKey();
    int i = 0;   // how far through buffer we are
    while (key != '#')
    {
        switch (key)
        {
            case NO_KEY:
                break;

            case '0': case '1': case '2': case '3': case '4':
            case '5': case '6': case '7': case '8': case '9':
                lcd.print(key);
                if (i < maxLength)
                {
                    result [i++] = key;
                    result [i] = 0;  // terminating null
                }
                break;

            case '*':  // backspace
                if (i > 0)
                {
                    i--;
                    result [i] = 0;  // terminating null
                }
                break;
        }

        key = kpd.getKey();
    }
} // end of GetNumber

I changed it..it gave me this error

Answer 1

 long GetNumber()

A long type (on this platform) can go up to 2^31 (that is 2,147,483,648). Large 10-digit numbers (ie. starting with "9") will not fit.

You could work with a 10-character string (or longer) instead of a long.

---

Make GetNumber deal with strings, like this:

void GetNumber(char * result, const int maxLength)
{
  char key = kpd.getKey();
  int i = 0;   // how far through buffer we are
  while (key != '#')
  {
    switch (key)
    {
      case NO_KEY:
        break;

      case '0' ... '9': 
        lcd.print(key);
        if (i < maxLength)
          {
          result [i++] = key;
          result [i] = 0;  // terminating null
          }
        break;

      case '*':  // backspace
        if (i > 0)
          {
          i--;
          result [i] = 0;  // terminating null
          }

        break;
    }

    key = kpd.getKey();
  }
} // end of GetNumber

---

Now, to call GetNumber supply a char buffer and the maximum length (allowing for the 0x00 needed to terminate a string), like this:

  char buf [15];
  GetNumber (buf, sizeof (buf) - 1);

Answer 2

Use a string or a char. They store their numbers a character per character and not a huge number. A long can hold a max of around 2 000 000 000.

For example, 123 declared as an lon would be litteraly, mathematicly 123 (as 100 + 20 + 3). On the other side, a char will consider it as 1,2,3. It can store higher digits number.