Hi,
I'm assuming it's trying to write two 16 bit values to the variable B6. Will one overwrite the other or will they combine to make one 32 bit word?
No, it will write W1 (a word), then W2 (a word), then B6 (a byte) to the EEPROM (a total of 5 bytes) starting at location 0. To be honest I don't know if it writes the least or the most significant byte first but that's easy to find out. It doesn't really matter if you use READ with the WORD modifier since I'm sure it's been designed to match.

If memory_output4 is a WORD and you want to store it in EEPROM starting at location 3 then
Code:
WRITE 3, WORD memory_output4
Then, to read it back
Code:
READ 3, WORD memory_output6
[/code]

/Henrik.