Why eeprom_read_byte/eeprom_write_byte use uint8_t type addr

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Silver Markadmin
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Posted by admin: Fri. Sep 26, 2003 - 06:40 AM
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In WINAVR include file, eeprom function defined like bellow:

extern uint8_t eeprom_read_byte (const uint8_t *addr);
^^^^^
extern uint16_t eeprom_read_word (const uint16_t *addr);
^^^^^
extern void eeprom_write_byte (uint8_t *addr, uint8_t val);
^^^^^
extern void eeprom_write_word (uint16_t *addr, uint16_t val);
^^^^^
The questions are:
1) Why using 8bit addres while read or write a bye?
2) Can I just read or write ONE byte using 16 bit address? How can I do it?

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Tools, Compilers and General Programming
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julius
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Posted by julius: Fri. Sep 26, 2003 - 07:21 AM
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The size of a address (i.e. pointer in C) is always the address size of the particular processor architecture. That is 16 bits with an AVR.

uint8_t *addr means that the pointer points to an 8 bit variable, not that it is an 8 bit address.

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Silver Markadmin
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Posted by admin: Fri. Sep 26, 2003 - 08:32 AM
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Thank you, Julius

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vish2207
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Posted by vish2207: Thu. Oct 29, 2015 - 05:12 AM
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Removed spam.

Last Edited: Thu. Oct 29, 2015 - 06:55 AM
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