I found a strange mismatch in the reference manual for the OTP memory size:
Accordingly to the manual (RM0385, page 78) the start-address of the OTP memory begins at 0x1FF0 F000 and ends at address: 0x1FF0 F41F (in AXIM mode) and is declared as a memory area with the size of 1024 bytes:
But if I calculate 0x1FF0F41F - 0x1FF0F000 + 1 I get a total of 1056 bytes OTP memory?!
Same addresses are set in the latest CMSIS header V1.17.0 (https://raw.githubusercontent.com/STMicroelectronics/STM32CubeF7/master/Drivers/CMSIS/Device/ST/STM32F7xx/Include/stm32f746xx.h):
#define FLASH_OTP_BASE 0x1FF0F000UL
#define FLASH_OTP_END 0x1FF0F41FUL
Is this a typo in the manual or is my calculation wrong?
I don't think it's a typo, and your calculation is correct.
My explanation is as follows:
Chapter 3.6 of the same Reference Manual shows the organization of the one-time programmable (OTP) part of the OTP area, and explains:
The OTP area is divided into 16 OTP data blocks of 64 bytes and one lock OTP block of 16 bytes. The OTP data and lock blocks cannot be erased. The lock block contains 16 bytes LOCKBi (0 ≤ i ≤ 15) to lock the corresponding OTP data block (blocks 0 to 15). (...)
So it consists of 1024 data bytes and 16 lock bytes. I guess the lock bytes are implemented as special "registers" and are not considered part of the sector. It's more a matter of definition and does not really matter that much.
There probably is a 1024-byte sector to write the (one-time programmable) data to, but the address range is slightly larger because the lock bits have to be addressed as well.
Related
I was trying to add a few system calls to the xv6 source code developed at MIT, and upon reading this resource (https://pdos.csail.mit.edu/6.S081/2020/xv6/book-riscv-rev1.pdf), on page 26, they consider the maximum possible virtual address that XV6RISC-V supports.
When I tried to look at the source code for the definition of MAXVA (MAXimum Virtual Address), I came across the following code snippet.
// one beyond the highest possible virtual address.
// MAXVA is actually one bit less than the max allowed by
// Sv39, to avoid having to sign-extend virtual addresses
// that have the high bit set.
#define MAXVA (1L << (9 + 9 + 9 + 12 - 1))
I was intrigued by the expression '9 + 9 + 9 + 12 - 1' instead of simply writing '38'. I tried to look up the underlying reasoning for this but did not find anything related. Is this some kind of optimization? If so, at what level is this relevant and where else could this be relevant?
I have some experience in assembly language programming and understand the basics of how the C code written is being translated to the assembly structure and how the final assembly code corresponding to bit shifting might look like, (using x86 salq and sarq or RISC-V slli). Any hints/ thoughts would be appreciated as well.
The answer could be guessed from vm.c
// The risc-v Sv39 scheme has three levels of page-table
// pages. A page-table page contains 512 64-bit PTEs.
// A 64-bit virtual address is split into five fields:
// 39..63 -- must be zero.
// 30..38 -- 9 bits of level-2 index.
// 21..29 -- 9 bits of level-1 index.
// 12..20 -- 9 bits of level-0 index.
// 0..11 -- 12 bits of byte offset within the page.
9+9+9 correspond to each 9 bits index to address the page
12 correspond to the offset in the page
For the compiler, write 9+9+9+12-1 is the same than 38, but for the human -aware of bits fields- reading this, the first is more clear.
I am using C_Decrypt with the CKM_AES_CBC_PAD mechanism. I know that my ciphertext which is 272 bytes long should actually decrypt to 256 bytes, which means a full block of padding was added.
I know that according to the standard when invoking C_Decrypt with a NULL output buffer the function may return an output length which is somewhat longer than the actual required length, in particular when padding is used this is understandable, as the function can't know how many padding bytes are in the final block without carrying out the actual decryption.
So the question is whether if I know that I should get exactly 256 bytes back, such as in the scenario I explained above, does it make sense that I am still getting a CKR_BUFFER_TOO_SMALL error as a result, despite passing a 256 bytes buffer? (To make it clear: I am indicating that this is the length of the output buffer in the appropriate output buffer length parameter, see the parameters of C_Decrypt to observe what I mean)
I am encountering this behavior with a Safenet Luna device and am not sure what to make of it. Is it my code's fault for not querying for the length first by passing NULL in the output buffer, or is this a bug on the HSM/PKCS11 library side?
One more thing I should perhaps mention is that when I provide a 272 (256+16) bytes output buffer, the call succeeds and I am noticing that I am getting back my expected plaintext, but also the padding block which means 16 final bytes with the value 0x10. However, the output length is updated correctly to 256, not 272 - this also proves that I am not using CKM_AES_CBC instead of CKM_AES_CBC_PAD accidentally, which I suspected for a moment as well :)
I have used CKM.AES_CBC_PAD padding mechanism with C_Decrypt in past. You have to make 2 calls to C_Decrypt (1st ==> To get the size of the plain text, 2nd ==> Actual decryption). see the documentation here which talks about determining the length of the buffer needed to hold the plain-text.
Below is the step-by-step code to show the behavior of decryption:
//Defining the decryption mechanism
CK_MECHANISM mechanism = new CK_MECHANISM(CKM.AES_CBC_PAD);
//Initialize to zero -> variable to hold size of plain text
LongRef lRefDec = new LongRef();
// Get ready to decrypt
CryptokiEx.C_DecryptInit(session_1, mechanism, key_handleId_in_hsm);
// Get the size of the plain text -> 1st call to decrypt
CryptokiEx.C_Decrypt(session_1, your_cipher, your_cipher.length, null, lRefDec);
// Allocate space to the buffer to store plain text.
byte[] clearText = new byte[(int)lRefDec.value];
// Actual decryption -> 2nd call to decrypt
CryptokiEx.C_Decrypt(session_1, eFileCipher, eFileCipher.length, eFileInClear,lRefDec);
Sometimes, decryption fails because your input encryption data was misleading (however, encryption is successful but corresponding decryption will fail) the decryption algorithm. So it is important not to send raw bytes directly to the encryption algorithm; rather encoding the input data with UTF-8/16 schema's preserves the data from getting misunderstood as network control bytes.
I'm using AT24C512 EEPROM which is 512KB along with my STM32
I'm able to write 128bytes of data at once using
HAL_I2C_Mem_Write(&_EEPROM24XX_I2C,0xa0,Address,I2C_MEMADD_SIZE_16BIT,(uint8_t*)data,size_of_data,100)
but the issue is that i want to write more data after the data that was just wrote, but the EEPROM will replace the data as the Address is the same
so how can i skip the written address ?
This answer is not about using HAL with I2C, but hope it will point you
Just check datasheet (I looking into STM32F0) and you can see that the limit is 255 bytes (register CR2:NBYTES), I'm not sure if there is another limitation in HAL, but using direct access to registers you can sent 255 bytes at once or fragment it and sent how much you want.
For fragmenting there is bit CR2:RELOAD, if you set this, then at the end will be not transfer stopped, and you can update next NBYTES, .. when you will set last block of bytes (which will fit into NBYTES) then clear bit CR2:RELOAD.
This has one disadvantage, that every 255 bytes, you will be interrupted.
i think you should check the AT24C512 datasheet page 7.
If more
than 128 data words are transmitted to the EEPROM, the
data word address will
“
roll over
”
and previous data will be
overwritten. The address
“
roll over
”
during write is from the
last byte of the current page to the first byte of the same
page.
I have C program running on a AVR32 microcontroller (UC3C0512C).
Issuing the avr32-size -A PROGRAM.elf command generates the following output:
PROGRAM.elf :
section size addr
.reset 8200 2147483648
.rela.got 0 2147491848
.text 99512 2147491848
.exception 512 2147591680
.rodata 5072 2147592192
.dalign 4 4
.data 7036 8
.balign 4 7044
.bss 5856 7048
.heap 48536 12904
.comment 48 0
.debug_aranges 8672 0
.debug_pubnames 14476 0
.debug_info 311236 0
.debug_abbrev 49205 0
.debug_line 208324 0
.debug_frame 23380 0
.debug_str 43961 0
.debug_loc 63619 0
.debug_macinfo 94469328 0
.stack 4096 61440
.data_hram0 512 2684354560
.debug_ranges 8368 0
Total 95379957
Can someone explain how to interpret these values?
How can I calculate the flash and ram usage based on this list?
Update 1:
Without the -A flag, I am getting the following:
text data bss dec hex filename
113296 7548 58496 179340 2bc8c PROGRAM.elf
Update 2:
I'm not using dynamic memory allocation, so according the avr-libc user-manual, the free RAM space should be simply: stackpointer minus __heap_start.
In this case: 61440 - 12904 = 48536 byte free RAM space.
Can someone confirm that?
(There is a mismatch in the two outputs in your question. The bss number is wildly different.)
If you don't use malloc, and don't count the stack, then yes, the RAM usage is the data plus the bss (plus some alignment spacing). The data are the variables that are set in a declaration, and the bss are the variables that are not. The C runtime will probably initialize them to 0, but it doesn't have to.
The flash usage will be the text and the data. That is, the flash will include the program instructions and C runtime, but also the values that need to get copied into RAM on startup to initialize those variables. This data is generally tacked onto the end of the program instructions.
Re: update 2
RAM holds global variables, the heap, and then the stack in that order.
The global variables can be initialized in the program, or not. The .data section is stored in flash, and the C runtime copies these values into the beginning of RAM where the corresponding variables live before your code runs. The .bss section of global variables needs space in RAM to hold the values, but they aren't necessarily initialized. The C runtime that comes with avr-gcc does actually initialize these to 0. The point it that your don't need to store an array of 0s to copy over, as you do with the .data section.
You are not using heap, but dynamically allocated memory is obtained from the addresses between heap_start and heap_end.
But the stack is not limited. Yes, the stack-pointer is initialized at startup, but it changes as your program runs, and can move well into the heap or even into the global variables (stack overflow). The stack pointer moves whenever a function is called, or local variables within a function are used. For example, a large array declared inside a function will go on the stack.
So in answer to your question, there is no RAM that is guaranteed to remain free.
I think you should remove the -A (all) flag, since that gives you the more low-level list you're showing.
The default output is easier to parse, and seems to directly state the values you're after.
Note: I didn't try this, not a system with an AVR toolchain installed.
I guess that in your linker script you have RAM at 0, and Flash at 0x80000000, so all things that need to go to RAM are at addresses 0+ (.stack is the last at 61440 (spanning next 4k)). So you would need a bit more that 64k of RAM. Everything else you have is flash.
That is provided that your linker script is correct.
Also see unwind's comment.
These values are the assembly language sections of the compiled C code. See the docs for the details. This article is also helpful.
The section titled .text represents the instruction section, i.e. the assembly instructions. The .data section represents the size of the variables (ints, arrays, etc.). The size column has the significant info, and it has the size of each section in bytes. The .stack and .heap represent the memory allocated in preparation for the execution of the program to set up the virtual memory.
You can try
avr-nm --print-size --radix d --demangle x.elf
to get the sizes in decimal notation.
Then you can copy & paste into a spreadsheet, filter, sort by the sections, and sum it up.
While reading data in socket its important either keep a message terminator symbol or add the Packet size information at the begening of the message.
If a terminator symbol is used and a binary message is sent there is no guarantee that the terminator symbol would not appear in the middle of the message (unless some special encoding is used).
On the other hand if size information is attached. size information is unsigned and if one byte is used for it it cannot be used to transfer messages longer than 256 bytes. if 4 byte integer is used. its not even guaranteed that 4 bytes will come a s whole. just 2 bytes of the size information may come can assuming the size information has arrived it may use that 2 bytes and rest of the integer data will be discarded. waiting for 4 bytes to be available on read buffer may cause infinite awaiting if only 3 bytes are available on the buffer (e.g. if total buffer is 7 bytes or 4077 bytes long).
here comes two possible ways
sizeInfo separator chunk
read until the separator is found once found read until sizeInfo bytes passed
keep an unreadyBytes initialized at 4 upon receiving the sizeInfo change it accordingly
which one of these two is safer to use ? Please Criticize
Edit
My central question is how to make sure that the size bytes has arrived properly. assuming messages are of variable size.
its not even guaranteed that 4 bytes will come a s whole. just 2 bytes of the size information may come can assuming the size information has arrived it may use that 2 bytes and rest of the integer data will be discarded. waiting for 4 bytes to be available on read buffer may cause infinite awaiting if only 3 bytes are available on the buffer (e.g. if total buffer is 7 bytes or 4077 bytes long).
If you have a 4 bytes length descriptor you should always read at least 4 bytes, because the sender should have written this bytes in every message your server is receiving. If you can't get them, maybe there has been a problem in transmission. I really can't understand your problem.
Anyway I'll suggest to you not to use any separator chunk.
Put an header at data blocks you are transmitting and use a buffer to reconstruct the packet flow.
You must at least read the header of a packet to determine its length.
You can define a basic structure for a packet:
struct packet{
uint32 id;
char payload[MAX_PAYLOAD_SIZE];
};
The you read data from socket storing them into a buffer:
struct packet buffer;
Then you can read the data from the socket:
int n;
n = read(newsocket, &buffer, sizeof(uint32) + MAX_PAYLOAD_SIZE);
read returns the number of bytes read. If you read exactly a packet from the sender, then n = id. Otherwise maybe you read more data (es. the sender sent to you more packets). If you are receiving a stream of data split into unit (represented by packet structures), then you may use an array of packet to store the complete packet received and a temporarily buffer to manage incoming fragments.
Something like:
struct packet buffer[MAX_PACKET_STORED];
char temp_buffer[MAX_PAYLOAD_SIZE + 4];
int n;
n = read(newsocket, &buffer, sizeof(uint32) + MAX_PAYLOAD_SIZE);
//here suppose have received a packet of 100 Byte payload + 32 bit of length + 100 Byte
//fragments of the next packet.
//then:
int first_pack_len, second_pack_len;
first_pack_len = *((uint32 *)&temp_buffer[0]); //retrieve packet length
memcpy(&packet_buffer[0], temp_buffer, first_pack_len + sizeof(uint32)) //store the first packet into the array
second_pack_data_available_in_buffer = n - (first_pack_len + sizeof(uint32)); //total bytes read minus length of the first packet read
second_pack_len = *((int *)&temp_buffer[first_pack_len + sizeof(uint32)]);
I hope to have been clear enough. But maybe I'm misunderstanding your question.
Pay attention also that if the 2 end-systems communicating could have different endiannes, so it's a better idea use htonl/ntohl function on length when sending/receving length value. But this is another issue)