writing half word 16 bits on flash stm32f429ZI

TARHAN SAMAH · ‎2018-05-23

Posted on May 23, 2018 at 10:05

Hello evry body , i can write a byte on flash of stm32f429ZI WÄ°TH success . for info i use nucleo 144 and sector 23 the last sector .

for example :

with these adresses

uint32_t locip1_adress=0x081E0001;//LAST sector for STM32429ZI sector 23

uint32_t locip2_adress=0x081E0002;//LAST sector for STM32429ZI sector 23

uint32_t locip3_adress=0x081E0003;//LAST sector for STM32429ZI sector 23

uint32_t locip4_adress=0x081E0004;//LAST sector for STM32429ZI sector 23

uint32_t remip1_adress=0x081E0005;//LAST sector for STM32429ZI sector 23

uint32_t remip2_adress=0x081E0006;//LAST sector for STM32429ZI sector 23

uint32_t remip3_adress=0x081E0007;//LAST sector for STM32429ZI sector 23

uint32_t remip4_adress=0x081E0008;//LAST sector for STM32429ZI sector 23

uint32_t ROV_min_adress=0x081E0009;//LAST sector for STM32429ZI sector 23 this data needs two bytes

uint32_t ROV_max_adress=0x081E00B;//LAST sector for STM32429ZI sector 23 this data needs two bytes

/* Unlock the Flash to enable the flash control register access *************/

HAL_FLASH_Unlock();

/* Fill EraseInit structure*/

EraseInitStruct.TypeErase = FLASH_TYPEERASE_SECTORS;

EraseInitStruct.VoltageRange = FLASH_VOLTAGE_RANGE_3;

EraseInitStruct.Sector = FLASH_SECTOR_23;

EraseInitStruct.NbSectors = 1;

//you need to erase entire sector before write anything

HAL_FLASHEx_Erase(&EraseInitStruct, &SECTORError);//we need to erase the sector before write

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, locip1_adress,locip1);//write one byte

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, locip2_adress,locip2);//write one byte

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, locip3_adress,locip3);//write one byte

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, locip4_adress,locip4);//write one byte

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, remip1_adress,remip1);//write one byte// we need to write also all other parameter because we delete all sector 23

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, remip2_adress,remip2);//write one byte

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, remip3_adress,remip3);//write one byte

HAL_FLASH_Program(FLASH_TYPEPROGRAM_BYTE, remip4_adress,remip4);//write one byte

HAL_FLASH_Program(FLASH_TYPEPROGRAM_HALFWORD,ROV_min_adress,Rover_ID_min[0]);//write one 16 bit

HAL_FLASH_Program(FLASH_TYPEPROGRAM_HALFWORD,ROV_max_adress,Rover_ID_max[0]);//write one 16 bit

HAL_FLASH_Lock();

But in THE HALFWORD writing i have prblm with 16 bit writing looks these adresses are not accepted

uint32_t ROV_min_adress=0x081E0009; ----> in my idea for one word ill use 0x081E0009+ 0x081E000A Two bytes --->is it right

uint32_t ROV_max_adress=0x081E00B;-----> WÄ°LL use 0x081E000B+0x081E000C two bytes

but the data at the reading of ROV_max_adress are wrong strangely ??

how should i select the adresses separation between them for writing half word can someone give me an example . thanks

could not find detail about mem data adresses in detail .

Tilen MAJERLE · ‎2018-05-24

Posted on May 24, 2018 at 14:39

Hello

doumandjisamah

‌,

To know when you have 16-bit aligned address, you can check LSB bit which must be 0, thus:

Every address is byte aligned, no matter on LSB bit value

Every second address is 16-bit (half-word) aligned
- 0 = 0b000
  0
  , 2 = 0b001
  0
  , 4 = 010
  0
  , 6 = 0b011
  0

Every 4th address is 32-bit (word) aligned, LSB and LSB-1 bits must be 0.
- 0 = 0b00
  00
  ;4 = 0b01
  00
  ;8 = 0b10
  00

In your case, address

0x081E0009

is odd,

LSB =1 => not 16-bit aligned, while

0x081E000A

or

0x081E000C

is even number,

LSB = 0 => 16-bit aligned.

If we go back to flash write process:

To be able to write in BYTE mode, you may use any valid flash address (LSB don't care)

To be able to write in HALF-WORD mode, your address must be 16-bits aligned (LSB must be 0)

To be able to write in WORD mode, your address must be 32-bits aligned (LSB and LSB - 1 must be 0)

To be able to write in QWORD mode, your address must be 64-bits aligned (LSB, LSB - 1 and LSB - 2 must be 0).

Hope it is clear now.

Best regards,

Tilen

View solution in original post

Tilen MAJERLE · ‎2018-05-23

Posted on May 23, 2018 at 10:22

Dear

‌,

please check FLASH_SR register for possible errors when programming.

Address must be 16-bits aligned, which is not your case.

uint32_t ROV_min_adress=0x081E0009;�?

should be

uint32_t ROV_min_adress=0x081E0008; //Either 8 and then it will use 8 + 9 bytesuint32_t ROV_min_adress=0x081E000A; //Or A and then it will use A + B bytes�?�?

Best regards,

Tilen

TARHAN SAMAH · ‎2018-05-23

Posted on May 23, 2018 at 13:23

Hello Mister Tilen , thanks for answering but i think you didnt understand my question ,

i have two datas to save for example:

Rover_ID_min[0]=0x0012;

Rover_ID_max[0]=

0x0044;

in these adresses

uint32_t ROV_min_adress=0x081E0009;

uint32_t ROV_min_adress=0x081E000B;

HAL_FLASH_Program(FLASH_TYPEPROGRAM_HALFWORD,ROV_min_adress,Rover_ID_min[0]);//write one 16 bit

,------> Ä°LL use 9 and A

HAL_FLASH_Program(FLASH_TYPEPROGRAM_HALFWORD,ROV_max_adress,Rover_ID_max[0]);//write one 16 bit

----->Ä°LL USE B AND C

SO WHY THERE Ä°S A MÄ°STAKE ON MY READED DATA since i separate wel the adresses

Tilen MAJERLE · ‎2018-05-23

Posted on May 23, 2018 at 13:52

Dear

doumandjisamah

‌,

what are the values you read back in this case?

Please note, addresses0x081E0009 and0x081E000Bare not 16-bits aligned and you cannot program to these addresses as 16-bits (HALFWORD). You may use addresses0x081E0008

and

0x081E000A or

0x081E000A

and

0x081E000C or use byte programming width.

Best regards,

Tilen

TARHAN SAMAH · ‎2018-05-23

Posted on May 23, 2018 at 14:36

i read diferent values from what i wrote ;

what is the meaning of 16 bits alligned where can i see the memory map details ? to understand how to select my adresses i just can see on memory map sector 23 0x081E 0000 -0X081F FFFF

Ä°F Ä° SELECT

uint32_t ROV_min_adress=0x081E0010;

uint32_t ROV_max_adress=0x081E0012;

Ä°T WORKS i just want the explanation about the adresses

TARHAN SAMAH · ‎2018-05-23

Posted on May 23, 2018 at 14:38

according to the doc RM 0090 PAGE 77

TARHAN SAMAH · ‎2018-05-23

Posted on May 23, 2018 at 15:23

Thanks much i got it a little thanks

TARHAN SAMAH · ‎2018-05-23

Posted on May 23, 2018 at 15:24

even levels of memory got it thanks

Tilen MAJERLE · ‎2018-05-24

Posted on May 24, 2018 at 14:39

Hello

doumandjisamah

‌,

To know when you have 16-bit aligned address, you can check LSB bit which must be 0, thus:

Every address is byte aligned, no matter on LSB bit value

Every second address is 16-bit (half-word) aligned
- 0 = 0b000
  0
  , 2 = 0b001
  0
  , 4 = 010
  0
  , 6 = 0b011
  0

Every 4th address is 32-bit (word) aligned, LSB and LSB-1 bits must be 0.
- 0 = 0b00
  00
  ;4 = 0b01
  00
  ;8 = 0b10
  00

In your case, address

0x081E0009

is odd,

LSB =1 => not 16-bit aligned, while

0x081E000A

or

0x081E000C

is even number,

LSB = 0 => 16-bit aligned.

If we go back to flash write process:

To be able to write in BYTE mode, you may use any valid flash address (LSB don't care)

To be able to write in HALF-WORD mode, your address must be 16-bits aligned (LSB must be 0)

To be able to write in WORD mode, your address must be 32-bits aligned (LSB and LSB - 1 must be 0)

To be able to write in QWORD mode, your address must be 64-bits aligned (LSB, LSB - 1 and LSB - 2 must be 0).

Hope it is clear now.

Best regards,

Tilen

TARHAN SAMAH · ‎2018-05-24

Posted on May 24, 2018 at 14:47

much much cleaaaaaaaaaaaaaar thanks much