Erasing EEPROM taking over 30 seconds

[mayurharge]

I have an EEPROM reset function where I erase EEPROM completely.
the function is something like this.

here EEPROM Size is 1024

{ DBG("erasing EEPROM.");
unsigned long start = millis();
for (unsigned int i = 0; i < EEPROM_SIZE; i++)
{
EEPROM.write(i, 0x00);
IWatchdog.reload();
}
DBG("Time took : ", millis() - start);
DBG("Erased eeprom");
// NVIC_SystemReset();
}

[602487] erasing EEPROM.

[632579] Time took : 30092

[632579] Erased eeprom.

to execute this function it's taking more than 30 seconds.
please look into this.

[fpistm]

Hi @mayurharge
could you be more precise.
Which board, host OS, IDE version,...

For reference, you should use EEPROM buffered access:
#302

Here example:
https://github.com/stm32duino/STM32Examples/blob/master/examples/NonReg/BufferedEEPROM/BufferedEEPROM.ino

[mayurharge]

Hi @mayurharge
could you be more precise.
Which board, host OS, IDE version,...

For reference, you should use EEPROM buffered access:
#302

Here example:
https://github.com/stm32duino/STM32Examples/blob/master/examples/NonReg/BufferedEEPROM/BufferedEEPROM.ino

Sorry for incomplete details, here is specific info

• Arduino IDE - 1.8.0
• OS - tested on both win10 & Mac OS
• Board - STM32F1 series (STM32F103Bt6 128kb version)

[fpistm]

Thanks for info.

As mentioned in #302 by @hasenbanck

The current eeprom emulation didn't had any buffered access to the
data on the flash. Every access to a byte would trigger a whole
page read / write. For writes it would mean, that the whole page
would be erased. For backward compatibility we will keep
the current functions as they are, but will add functions
to fill/flush the buffer and read from it.

This means with your code, a loop of 1024, will erase and write 1024 times the 1kb flash page.
So, as a write takes 30ms you get 30s for the total EEPROM size.

To avoid this @hasenbanck provides a way to work on buffered EEPROM and when need write only once:

// Clear the EEPROM buffer
  for (uint16_t i = 0; i < EEPROM.length(); i++) {
    eeprom_buffered_write_byte(i, 0);
  }
  // Copy the data from the buffer to the flash
  eeprom_buffer_flush();

In this case this takes only 30ms to clear the flash.

[mayurharge]

Thanks for info.

As mentioned in #302 by @hasenbanck

The current eeprom emulation didn't had any buffered access to the
data on the flash. Every access to a byte would trigger a whole
page read / write. For writes it would mean, that the whole page
would be erased. For backward compatibility we will keep
the current functions as they are, but will add functions
to fill/flush the buffer and read from it.

This means with your code, a loop of 1024, will erase and write 1024 times the 1kb flash page.
So, as a write takes 30ms you get 30s for the total EEPROM size.

To avoid this @hasenbanck provides a way to work on buffered EEPROM and when need write only once:

// Clear the EEPROM buffer
  for (uint16_t i = 0; i < EEPROM.length(); i++) {
    eeprom_buffered_write_byte(i, 0);
  }
  // Copy the data from the buffer to the flash
  eeprom_buffer_flush();

In this case this takes only 30ms to clear the flash.

Thanks for the suggestion, implemented the same in my code.