On 12 Dec 2017 at 11:31, Isaac M. Bavaresco wrote:

> The mechanism of the sequence count works differently than the time
> roll-over.
>=20
> With two bytes you will have 65536 possible values, but your EEPROM
> memory has room for, say, 128 records.
>=20
> So if any record has sequence number more than 128 counts AHEAD of the
> one you have, it is older. We could increase (double it, for instance)
> this threshold to account for some lost record in-between.
>=20
> You compare sequence numbers by subtracting them and doing an unsigned
> compare:
>=20
> if( (uint16_t)( Seq2 - Seq1 ) < THRESHOLD )
> =A0=A0=A0 {
> =A0=A0=A0 /* Record with sequence Seq2 is newer than record with sequence=
 Seq1 */
> =A0=A0=A0 }
>=20
> This method takes care of roll-over, you will never need to check for
> overflow.
>=20
> Example:
>=20
> #define THRESHOLD=A0=A0=A0 256
>=20
> int=A0=A0=A0 NewestRecord=A0=A0=A0 =3D -1;
> for( int i =3D 0; i < NumRecords; i++ )
> =A0=A0=A0 {
> =A0=A0=A0 if( ValidateRecord( i ))
> =A0=A0=A0 =A0=A0=A0 {
> =A0=A0=A0 =A0=A0=A0 if( NewestRecord < 0 || (uint16_t)( GetSequence( i ) =
-
> GetSequence( NewestRecord )) < THRESHOLD )
> =A0=A0=A0 =A0=A0=A0 =A0=A0=A0 NewestRecord=A0=A0=A0 =3D i;
> =A0=A0=A0 =A0=A0=A0 }
> =A0=A0=A0 }
>=20
> if( NewestRecord < 0 )
> =A0=A0=A0 {
> =A0=A0=A0 /* There is no valid record. */
> =A0=A0=A0 }
> else
> =A0=A0=A0 {
> =A0=A0=A0 /* This is the record with the latest saved value. */
> =A0=A0=A0 }
>=20

Ok, cool, I see how that works, nice.

For that matter though it is conceivable that the record counter and the ho=
urs=20
counter could be the same thing. Kind of what I was thinking all along but =
probably=20
never spelled it out. Count increments by a value of one each time interval=
, store=20
record in next EEPROM location(s). At power up find the highest record=20
number/hours count using the method you descibe.

The only slight challenge I will have, and not mentioned previously, is tha=
t a service=20
person has access to change the engine hours count - for example when servi=
cing=20
a machine and swapping out this module. In that one and only case the "new"=
 hours=20
count could be lower than any of the "old" records stored in EEPROM, or "ah=
ead"=20
by more than the number of records stored. The solution would be to delete,=
 or=20
mark as invalid, all the old values at that time. I don't like to have code=
 that does=20
this, with the chance that it could run errantly, but concede it may be nec=
essary. I=20
have safeguards in place to reduce the chances of the service person routin=
es=20
being accessed improperly.
--=20
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