# Registers and bits? I dunno know LOL

Im trying to understand what does it A0-A8 mean when accessing an EEPROM that has 512x8bits of memory.

Does that mean that A0=0b00000000 has 512 bits of information? If so why are there 9 Adresses (A0-A8) that makes 9x512 more than 4096.

A0 to A7 = 8 address bits, 256 addresses
adding A8 doubles that, to 512 addresses
Address range is then 0x0000 to 0x01FF
8 bits x 512 addresses = 4096 bits.

LOL. I'm an idiot. I understand now.

The memory address has 9 bits because it represents 512. and each address has 8bits, giving me 4096bits.

So A0-A8 in the datasheet means that the memory address must have nine digits (or bits), from 0b000000000 to 0b111111111.

I get it now.

Another question.

So, if the datasheet says that for reading a location the address must be A8-A0 (is there a differrence if it said A0-A8?), then that means that when I bit bang a 1 on the 1st bit then shiftout(0b00000000) does that mean the eeprom reads it as 0b000000001 or 0b100000000?

Post a link to the datasheet. Would just be guessing otherwise.