EEPROM.read/write bug

Using Arduino Programming Questions
1 
I am trying to save long numbers to eeprom when I push a button, but when I display the number, it just shows a number around 29149, every time. I have tried using different boards, nano and uno, and have the same result. Here is my code, any suggestions?


#include <EEPROM.h>
int button1 = 8;
int a1 = EEPROM.read(18);
int b1 = EEPROM.read(17);
int c1 = EEPROM.read(16);
int d1 = EEPROM.read(15);
int e1 = EEPROM.read(14);
int f1 = EEPROM.read(13);
int g1 = EEPROM.read(12);
int h1 = EEPROM.read(11);
int i1 = EEPROM.read(10);
int j1 = EEPROM.read(9);
int k1 = EEPROM.read(8);
int l1 = EEPROM.read(7);
int m1 = EEPROM.read(6);
int o1 = EEPROM.read(5);
int p1 = EEPROM.read(4);
int q1 = EEPROM.read(3);
int r1 = EEPROM.read(2);
int s1 = EEPROM.read(1);
int t1 = EEPROM.read(0);

int y =  ((100000000000000000+s1)+(10000000000000000+r1)+(1000000000000000+q1)
+(100000000000000+p1)+(10000000000000+o1)+(1000000000000+m1)+(100000000000+l1)+(10000000000+k1)
+(1000000000+j1)+(100000000+i1)+(10000000+h1)+(1000000+g1)+(100000+f1)+(10000+e1)+(1000+d1)+(100+c1)+(10+b1)+a1);
int x = y;
void setup(){
  Serial.begin(9600);
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();
  Serial.println();

  Serial.println("Solving...");
pinMode(button1, INPUT_PULLUP);
}


void loop(){
  if(x % 2 == 0){
      x = x/2;
  }
  else{
      x=3*x;
      x++;
  }
if(x<y){
Serial.println(y);
y++;
 x=y;

}
if(digitalRead(button1) == false){
  
  unsigned n = y;
  Serial.println("Saving...");
  EEPROM.write(18, (n / 1U) % 10);
  EEPROM.write(17, (n / 10U) % 10);
  EEPROM.write(16, (n / 100U) % 10);
  EEPROM.write(15, (n / 1000U) % 10);
  EEPROM.write(14, (n / 10000U) % 10);
  EEPROM.write(13, (n / 100000U) % 10);
  EEPROM.write(12, (n / 1000000U) % 10);
  EEPROM.write(11, (n / 10000000U) % 10);
  EEPROM.write(10, (n / 100000000U) % 10);
  EEPROM.write(9,  (n / 1000000000U) % 10);
  EEPROM.write(8,  (n / 10000000000U) % 10);
  EEPROM.write(7,  (n / 100000000000U) % 10);
  EEPROM.write(6,  (n / 1000000000000U) % 10);
  EEPROM.write(5,  (n / 10000000000000U) % 10);
  EEPROM.write(4,  (n / 100000000000000U) % 10);
  EEPROM.write(3,  (n / 1000000000000000U) % 10);
  EEPROM.write(2,  (n / 10000000000000000U) % 10);
  EEPROM.write(2,  (n / 100000000000000000U) % 10);
  EEPROM.write(0,  (n / 1000000000000000000U) % 10);
delay(2000);                   
Serial.print("Saved at: ");
Serial.println(y);
delay(2000);                                         
}
}
2

use put() and get() and you can deal with your long directly

Do you expect this to be considered as a binary number?

it's not. and it's way too big for an int

3

1. What is the size of a long-type number in hex base and decimal base?
2. What is the size of a byte-type number in hex base and decimal base?
3. How many bytes of data are written into EEPROM when the following code is executed:

EEPROM.write(ar1, arg2);
4

Regardless of how you are reading the EEPROM, your method of reconstructing a long from individual bytes is flawed. Usually that would be done with bit shifting, or reading the individual bytes into a byte array, then memcpy into a long. (Possible to use a union, but that may be dependent on the compiler). Your method of saving the number to eeprom is also fairly unusual.

put and get are the easiest method.

5

This topic was automatically closed 180 days after the last reply. New replies are no longer allowed.