Re: [avr-gcc-list] read write eeprom WINAVR20050214

[E. Weddington]

Eivind Sivertsen wrote:

> > > The address is an address to an 8 bit variable (which is 16 bits on  
> > > AVR), not
> > > a 8 bit address. Basic C pointer stuff.
> > >
> > >      
> > ----------------------------------------------
> > Oh yes, you are right. silly me!
> >
> > --Royce.
> >    
>
>
> - But the value to write is a  uint16_t ...? Look: 
> void 	eeprom_write_word (uint16_t *addr, uint16_t val)
>
>
>  
Hi Eivind,

The functions in question are these:

uint8_t     eeprom_read_byte (const uint8_t *addr)
uint16_t     eeprom_read_word (const uint16_t *addr)
void     eeprom_write_byte (uint8_t *addr, uint8_t val)
void     eeprom_write_word (uint16_t *addr, uint16_t val)

The question had to do with eeprom_read_byte(). There are a lot of 
people who seem to "freak out" when they see that "uint8_t" in the 
parameter listing, and they think: "Oh, no! I can only write to the 
first 256 bytes in eeprom!"

But they don't look carefully enough to see that addr is a *pointer to 
an uint8_t*, and realize that a *pointer* is 16 bits. The pointer is 
pointing to a byte value, but the pointer itself can address up to 64K 
worth of memory.

This is a typical C language newbie question. And we should probably put 
this in the avr-libc FAQ or something.....

Eric