why can't program FLASH when FLASH all bit is 1?

Question

I use STM32L431, it only possible to program double word, I want program a new data to FLASH if the 64 bits all is '1' in FLASH, otherwise, if the 64 bits is not all '1', I erase it first before program. But when I program a 0xffffffffffffffff to the 64 bits and then need to program a new data(for example: 0x1122334455667788), it fail to program.void Flash_Test(void){	uint32_t startAddr = 0x0801C000;	HAL_StatusTypeDef status;	FLASH_EraseInitTypeDef EraseInitStruct;	uint32_t pageErr;​	printf("erase one page
");	/*erase one page*/	HAL_FLASH_Unlock();	__HAL_FLASH_CLEAR_FLAG(FLASH_FLAG_OPTVERR); 	EraseInitStruct.TypeErase  = FLASH_TYPEERASE_PAGES;	EraseInitStruct.Banks    = GetBank(startAddr);	EraseInitStruct.Page    = GetPage(startAddr);	EraseInitStruct.NbPages   = 1;	status = HAL_FLASHEx_Erase(&EraseInitStruct, &pageErr);	HAL_FLASH_Lock();	if (status != HAL_OK)		return;​	printf("before write 0xff to one page
");	for (int i = 0; i < 2048; i += 4)	{		printf("%x ", *(volatile uint32_t*)(startAddr + i));	}	printf("
");​	/*write 0xff to one page*/	HAL_FLASH_Unlock();	__HAL_FLASH_CLEAR_FLAG(FLASH_FLAG_OPTVERR);	for (int i = 0; i < 2048; i += 8)	{		status = HAL_FLASH_Program(FLASH_TYPEPROGRAM_DOUBLEWORD, startAddr + i, 0xffffffffffffffff);		if (status != HAL_OK)		{			printf("Fail to write all pages!
");			break;		}	}	HAL_FLASH_Lock();​	printf("after write 0xff to one page
");	for (int i = 0; i < 2048; i += 4)	{		printf("%x ", *(volatile uint32_t*)(startAddr + i));	}	printf("
");​	printf("write 0xff
");	/*write 8 bytes*/	HAL_FLASH_Unlock();	__HAL_FLASH_CLEAR_FLAG(FLASH_FLAG_OPTVERR);	status = HAL_FLASH_Program(FLASH_TYPEPROGRAM_DOUBLEWORD, startAddr, 0xffffffffffffffff);	if (status != HAL_OK)	{		printf("Fail to write 8 bytes!
");	}	HAL_FLASH_Lock();}Run this sample code, it will print "Fail to write 8 bytes".Possible to program 1 to 0 and can't program 0 to 1 is FLASH's feature. In this case, even if all the bits is '1' in FLASH, HAL_FLASH_Program still return HAL_ERROR. Why?Thanks!

GCord.1 · Accepted Answer

Zzhou.2:waclawek.jan's comment regarding the "hidden ECC bits getting programmed with a non-FF value" did not mean to imply that they were programmed incorrectly; indeed, the value programmed is correct -- that is, reading back the 0xFFFFFFFF will not result in a double-bit error (ECCD flag). What it means when the ECC bits are programmed to a non-FF value is that the corresponding STM32 Flash location is no longer alterable. In other words, when the device is erased, each 64-bits of Flash data and the associated 8 ECC bits all contain ''1' . This represents something of an "impossible" condition, since the ECC bits for a 64-bit value of "all FFs" is NOT itself 0xFF. You can perhaps understand this as being a specific condition rather than just one of many value settings. In any case, each of these '1's can be programmed to a '0' using the Flash write procedure, but any '0's cannot be programmed back to '1' -- they can only be erased to achieve that change. So, if you take a double-word Flash location that already contains 64 bits of '1's and perform a Flash write of (0xFFFFFFFF, 0xFFFFFFFF), it results in the same value but a different condition: that is, the 8 ECC bits will have changed. And once you've changed those, you cannot change them back. So, the writing of even a (0xFFFFFFFF,0xFFFFFFFF) will result in some '0' bits (in the corresponding ECC location), which will prevent you from being able to re-write that location. So, the error you should be getting when you try is not ECCD, but rather simply that the location could not be re-written.

Tesla DeLorean · Answer

You get to program the cells ONCE, for the 0xFFFFFFFFF, 0xFFFFFFFFF case opt not to write them the first time around

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