I have been attempting to retrieve ID3V2 Tag Frames by parsing through the mp3 file and retrieving each frame's size. So far I have had no luck.
I have effectively allocated memory to a buffer to aid in reading the file and have been successful in printing out the header version but am having difficulty in retrieving both the header and frame sizes. For the header framesize I get 1347687723, although viewing the file in a hex editor I see 05 2B 19.
Two snippets of my code:
typedef struct{ //typedef structure used to read tag information
char tagid[3]; //0-2 "ID3"
unsigned char tagversion; //3 $04
unsigned char tagsubversion;//4 00
unsigned char flags; //5-6 %abc0000
uint32_t size; //7-10 4 * %0xxxxxxx
}ID3TAG;
if(buff){
fseek(filename,0,SEEK_SET);
fread(&Tag, 1, sizeof(Tag),filename);
if(memcmp(Tag.tagid,"ID3", 3) == 0)
{
printf("ID3V2.%02x.%02x.%02x \nHeader Size:%lu\n",Tag.tagversion,
Tag.tagsubversion, Tag.flags ,Tag.size);
}
}
Due to memory alignment, the compiler has set 2 bytes of padding between flags and size. If your struct were putted directly in memory, size would be at address 6 (from the beginning of the struct). Since an element of 4 bytes size must be at an address multiple of 4, the compiler adds 2 bytes, so that size moves to the closest multiple of 4 address, which is here 8. So when you read from your file, size contains bytes 8-11. If you try to print *(&Tag.size - 2), you'll surely get the correct result.
To fix that, you can read fields one by one.
ID3v2 header structure is consistent across all ID3v2 versions (ID3v2.0, ID3v2.3 and ID3v2.4).
Its size is stored as a big-endian synch-safe int32
Synchsafe integers are
integers that keep its highest bit (bit 7) zeroed, making seven bits
out of eight available. Thus a 32 bit synchsafe integer can store 28
bits of information.
Example:
255 (%11111111) encoded as a 16 bit synchsafe integer is 383
(%00000001 01111111).
Source : http://id3.org/id3v2.4.0-structure § 6.2
Below is a straightforward, real-life C# implementation that you can easily adapt to C
public int DecodeSynchSafeInt32(byte[] bytes)
{
return
bytes[0] * 0x200000 + //2^21
bytes[1] * 0x4000 + //2^14
bytes[2] * 0x80 + //2^7
bytes[3];
}
=> Using values you read on your hex editor (00 05 EB 19), the actual tag size should be 112025 bytes.
By coincidence I am also working on an ID3V2 reader. The doc says that the size is encoded in four 7-bit bytes. So you need another step to convert the byte array into an integer... I don't think just reading those bytes as an int will work because of the null bit on top.
Related
I am trying to convert the input from a device (always integer between 1 and 600000) to four 8-bit integers.
For example,
If the input is 32700, I want 188 127 00 00.
I achieved this by using:
32700 % 256
32700 / 256
The above works till 32700. From 32800 onward, I start getting incorrect conversions.
I am totally new to this and would like some help to understand how this can be done properly.
Major edit following clarifications:
Given that someone has already mentioned the shift-and-mask approach (which is undeniably the right one), I'll give another approach, which, to be pedantic, is not portable, machine-dependent, and possibly exhibits undefined behavior. It is nevertheless a good learning exercise, IMO.
For various reasons, your computer represents integers as groups of 8-bit values (called bytes); note that, although extremely common, this is not always the case (see CHAR_BIT). For this reason, values that are represented using more than 8 bits use multiple bytes (hence those using a number of bits with is a multiple of 8). For a 32-bit value, you use 4 bytes and, in memory, those bytes always follow each other.
We call a pointer a value containing the address in memory of another value. In that context, a byte is defined as the smallest (in terms of bit count) value that can be referred to by a pointer. For example, your 32-bit value, covering 4 bytes, will have 4 "addressable" cells (one per byte) and its address is defined as the first of those addresses:
|==================|
| MEMORY | ADDRESS |
|========|=========|
| ... | x-1 | <== Pointer to byte before
|--------|---------|
| BYTE 0 | x | <== Pointer to first byte (also pointer to 32-bit value)
|--------|---------|
| BYTE 1 | x+1 | <== Pointer to second byte
|--------|---------|
| BYTE 2 | x+2 | <== Pointer to third byte
|--------|---------|
| BYTE 3 | x+3 | <== Pointer to fourth byte
|--------|---------|
| ... | x+4 | <== Pointer to byte after
|===================
So what you want to do (split the 32-bit word into 8-bits word) has already been done by your computer, as it is imposed onto it by its processor and/or memory architecture. To reap the benefits of this almost-coincidence, we are going to find where your 32-bit value is stored and read its memory byte-by-byte (instead of 32 bits at a time).
As all serious SO answers seem to do so, let me cite the Standard (ISO/IEC 9899:2018, 6.2.5-20) to define the last thing I need (emphasis mine):
Any number of derived types can be constructed from the object and function types, as follows:
An array type describes a contiguously allocated nonempty set of objects with a particular member object type, called the element type. [...] Array types are characterized by their element type and by the number of elements in the array. [...]
[...]
So, as elements in an array are defined to be contiguous, a 32-bit value in memory, on a machine with 8-bit bytes, really is nothing more, in its machine representation, than an array of 4 bytes!
Given a 32-bit signed value:
int32_t value;
its address is given by &value. Meanwhile, an array of 4 8-bit bytes may be represented by:
uint8_t arr[4];
notice that I use the unsigned variant because those bytes don't really represent a number per se so interpreting them as "signed" would not make sense. Now, a pointer-to-array-of-4-uint8_t is defined as:
uint8_t (*ptr)[4];
and if I assign the address of our 32-bit value to such an array, I will be able to index each byte individually, which means that I will be reading the byte directly, avoiding any pesky shifting-and-masking operations!
uint8_t (*bytes)[4] = (void *) &value;
I need to cast the pointer ("(void *)") because I can't bear that whining compiler &value's type is "pointer-to-int32_t" while I'm assigning it to a "pointer-to-array-of-4-uint8_t" and this type-mismatch is caught by the compiler and pedantically warned against by the Standard; this is a first warning that what we're doing is not ideal!
Finally, we can access each byte individually by reading it directly from memory through indexing: (*bytes)[n] reads the n-th byte of value!
To put it all together, given a send_can(uint8_t) function:
for (size_t i = 0; i < sizeof(*bytes); i++)
send_can((*bytes)[i]);
and, for testing purpose, we define:
void send_can(uint8_t b)
{
printf("%hhu\n", b);
}
which prints, on my machine, when value is 32700:
188
127
0
0
Lastly, this shows yet another reason why this method is platform-dependent: the order in which the bytes of the 32-bit word is stored isn't always what you would expect from a theoretical discussion of binary representation i.e:
byte 0 contains bits 31-24
byte 1 contains bits 23-16
byte 2 contains bits 15-8
byte 3 contains bits 7-0
actually, AFAIK, the C Language permits any of the 24 possibilities for ordering those 4 bytes (this is called endianness). Meanwhile, shifting and masking will always get you the n-th "logical" byte.
It really depends on how your architecture stores an int. For example
8 or 16 bit system short=16, int=16, long=32
32 bit system, short=16, int=32, long=32
64 bit system, short=16, int=32, long=64
This is not a hard and fast rule - you need to check your architecture first. There is also a long long but some compilers do not recognize it and the size varies according to architecture.
Some compilers have uint8_t etc defined so you can actually specify how many bits your number is instead of worrying about ints and longs.
Having said that you wish to convert a number into 4 8 bit ints. You could have something like
unsigned long x = 600000UL; // you need UL to indicate it is unsigned long
unsigned int b1 = (unsigned int)(x & 0xff);
unsigned int b2 = (unsigned int)(x >> 8) & 0xff;
unsigned int b3 = (unsigned int)(x >> 16) & 0xff;
unsigned int b4 = (unsigned int)(x >> 24);
Using shifts is a lot faster than multiplication, division or mod. This depends on the endianess you wish to achieve. You could reverse the assignments using b1 with the formula for b4 etc.
You could do some bit masking.
600000 is 0x927C0
600000 / (256 * 256) gets you the 9, no masking yet.
((600000 / 256) & (255 * 256)) >> 8 gets you the 0x27 == 39. Using a 8bit-shifted mask of 8 set bits (256 * 255) and a right shift by 8 bits, the >> 8, which would also be possible as another / 256.
600000 % 256 gets you the 0xC0 == 192 as you did it. Masking would be 600000 & 255.
I ended up doing this:
unsigned char bytes[4];
unsigned long n;
n = (unsigned long) sensore1 * 100;
bytes[0] = n & 0xFF;
bytes[1] = (n >> 8) & 0xFF;
bytes[2] = (n >> 16) & 0xFF;
bytes[3] = (n >> 24) & 0xFF;
CAN_WRITE(0x7FD,8,01,sizeof(n),bytes[0],bytes[1],bytes[2],bytes[3],07,255);
I have been in a similar kind of situation while packing and unpacking huge custom packets of data to be transmitted/received, I suggest you try below approach:
typedef union
{
uint32_t u4_input;
uint8_t u1_byte_arr[4];
}UN_COMMON_32BIT_TO_4X8BIT_CONVERTER;
UN_COMMON_32BIT_TO_4X8BIT_CONVERTER un_t_mode_reg;
un_t_mode_reg.u4_input = input;/*your 32 bit input*/
// 1st byte = un_t_mode_reg.u1_byte_arr[0];
// 2nd byte = un_t_mode_reg.u1_byte_arr[1];
// 3rd byte = un_t_mode_reg.u1_byte_arr[2];
// 4th byte = un_t_mode_reg.u1_byte_arr[3];
The largest positive value you can store in a 16-bit signed int is 32767. If you force a number bigger than that, you'll get a negative number as a result, hence unexpected values returned by % and /.
Use either unsigned 16-bit int for a range up to 65535 or a 32-bit integer type.
I'm struggling with a problem that requires I perform a hex dump to an object file I've created with the function fopen().
I've declared the necessary integer variable (in HEX) as follows:
//Declare variables
int code = 0xCADE;
The output must be big Endian so I've swapped the bytes in this manner:
//Swap bytes
int swapped = (code>>8) | (code<<8);
I then opened the file for binary output in this manner:
//Open file for binary writing
FILE *dest_file = fopen(filename, "wb");
Afterwards, I write the variable code (which corresponds to a 16 bit word) to the file in the following manner using fwrite():
//Write out first word of header (0xCADE) to file
fwrite(&swapped, sizeof(int), 1, dest_file);
After compiling, running, and performing a hexdump on the file in which the contents have been written to, I observe the following output:
0000000 ca de ca 00
0000004
Basically everything is correct up until the extra "ca 00". I am unsure why that is there and need it removed so that my output is just:
0000000 ca de
0000004
I know the Endianness problem has been addressed extensively on the stack, but after performing a serach, I am unclear as to how to classify this problem. How can I approach this problem so that "ca 00" is removed?
Thanks very much.
EDIT:
I've changed both:
//Declare variables
int code = 0xCADE;
//Swap bytes
int swapped = (code>>8) | (code<<8);
to:
//Declare variables
unsigned short int code = 0xCADE;
//Swap bytes
unsigned short int swapped = (code>>8) | (code<<8);
And I observe:
0000000 ca de 00 00
0000004
Which gets me closer to what I need but there's still that extra "00 00". Any help is appreciated!
You are telling fwrite to write sizeof(int) bytes, which on your system evaluates to 4 bytes (the size of int is 4). If you want to write two bytes, just do:
fwrite(&swapped, 2, 1, dest_file);
To reduce confusion, code that reorders bytes should use bytes (uint8 or char) and not multi-byte types like int.
To swap two bytes:
char bytes[2];
char temp;
fread(bytes, 2, 1, file1);
temp = bytes[0];
bytes[0] = bytes[1];
bytes[1] = temp;
fwrite(bytes, 2, 1, file2);
If you use int, you probably deceive yourself assuming that its size is 2 (while it's most likely 4), and assuming anything about how your system writes int to files, which may be incorrect. While if you work with bytes, there cannot be any surprises - your code does exactly what it looks like it does.
I've been trying to brush up on my C recently and was writing a program to manually parse through a PNG file.
I viewed the PNG file in a hex editor and noticed a stream of bytes that looked like
00 00 00 0D
in hex format.
This string supposedly represents a length that I am interested in.
I used getc(file) to pull in the bytes of the PNG file.
I created a char array as
char example[8];
to store the characters retrieved from getc.
Now, I have populated example and printing it with
printf("%#x, %#x, %#x, %#x", example[0]....
shows 0, 0, 0, 0xd which is exactly what I want.
However when I use
int x = atoi(example)
or
int x = strtol(example, NULL, 16)
I get back zero in both cases (I was expecting 13). Am I missing something fundamental?
atoi converts strings like "0" to its numeric equivalent, in this case 0. What you have instead is the string "\0\0\0\0\0\0\0\r" which is nowhere near numeric characters.
If you want to interpret your bytes as a number you could do something like
char example[4] = {0, 0, 0, 0xd};
printf("%d\n", *(uint32_t*) example);
You will notice (in case you're using a x86 CPU) that you will get 218103808 instead of 13
due to little endianness: the farther you go right the more significant the number gets.
As PNG uses big endian you can simply use be32toh (big endian to host endianess):
uint32_t* n = example;
printf("%u\n", be32toh(*n)
atoi and strtol expect text strings, while you have an array of binary values. To combine the individual bytes in an array to a larger integer, try something like:
uint32_t x = (a[0] << 24) | (a[1] << 16) | (a[2] << 8) | a[3];
atoi etc. operates on (ascii) strings.
You would get 123 for "123", which is in bytes 49 50 41 0.
What you have instead is binary 00 00 00 7B ... (well, endianess matters too).
Simple, but in this case wrong solution (ignoring endianess):
Cast the array address to int* and then get a value with *.
As integers in PNG are supposed to be big endian in any case,
the pointer casting would only work with big endian machines.
As portable solution, shifting the bytes with 24,16,8,0 and binary-or´ing them will do.
I read this code in a library which is used to display a bitmap (.bmp) to an LCD.
I do really hard in understanding what is happening at the following lines, and how it does happen.
Maybe someone can explain this to me.
uint16_t s, w, h;
uint8_t* buffer; // does get malloc'd
s = *((uint16_t*)&buffer[0]);
w = *((uint16_t*)&buffer[18]);
h = *((uint16_t*)&buffer[22]);
I guess it's not that hard for a real C programmer, but I am still learning, so I thought I just ask :)
As far as I understand this, it sticks somehow together two uint8_tvariables to an uint16_t.
Thanks in advance for your help here!
In the code you've provided, buffer (which is an array of bytes) is read, and values are extracted into s, w and h.
The (uint16_t*)&buffer[n] syntax means that you're extracting the address of the nth byte of buffer, and casting it into a uint16_t*. The casting tells the compiler to look at this address as if points at a uint16_t, i.e. a pair of uint8_ts.
The additional * in the code dereferences the pointer, i.e. extracts the value from this address. Since the address now points at a uint16_t, a uint16_t value is extracted.
As a result:
s gets the value of the first uint16_t, i.e. bytes 0 and 1.
w gets the value of the tenth uint16_t, i.e. bytes 18 and 19.
h gets the value of the twelveth uint16_t, i.e. bytes 22 and 23.
The code:
takes two bytes at positions 0 and 1 in the buffer, sticks them together into an unsigned 16-bit value, and stores the result in s;
it does the same with bytes 18/19, storing the result in w;
ditto for bytes 22/23 and h.
It is worth noting that the code uses the native endianness of the target platform to decide which of the two bytes represents the top 8 bits of the result, and which represents the bottom 8 bits.
uint8_t* buffer; // pointer to 8 bit or simply one byte
Buffer points to memory address of bytes -> |byte0|byte1|byte2|....
(uint16_t*)&buffer[0] // &buffer[0] is actually the same as buffer
(uint16_t*)&buffer[0] equals (uint16_t*)buffer; it points to 16 bit or halfword
(uint16_t*)buffer points to memory: |byte0byte1 = halfword0|byte2byte3 = halfword1|....
w = *((uint16_t*)&buffer[18]);
Takes memory address to byte 18 in buffer, then reinterpret this address to address of halfword then gets halfword on this address;
it's simply w = byte18 and byte19 sticked together forming a halfword
h = *((uint16_t*)&buffer[22]);
h = byte22 and byte 23 sticked together
UPD More detailed explanation:
h = *((uint16_t*)&buffer[22]) =>
1) buffer[22] === 22nd uint8_t (a.k.a. byte) of buffer; let's call it byte22
2) &buffer[22] === &byte === address of byte22 in memory; it's of type uint8_t*, as same as buffer; letscall it byte22_address;
3) (uint16_t*)&buffer[22] = (uint16_t*)byte22_address; casts address of byte to address of (two bytes sticked together; address of halfword of the same address; let's call it halfword11_address;
4) h = *((uint16_t*)&buffer[22]) === *halfword11_address; * operator takes value at address, that is 11th halfword or bytes 22 and 23 sticked together;
could somebody explain to me why in 24-bit rgb bitmap file I have to add a padding which size depends on width of image ? What for ?
I mean I must add this code to my program (in C):
if( read % 4 != 0 ) {
read = 4 - (read%4);
printf( "Padding: %d bytes\n", read );
fread( pixel, read, 1, inFile );
}
Because 24 bits is an odd number of bytes (3) and for a variety of reasons all the image rows are required to start at an address which is a multiple of 4 bytes.
According to Wikipedia, the bitmap file format specifies that:
The bits representing the bitmap pixels are packed in rows. The size of each row is rounded up to a multiple of 4 bytes (a 32-bit DWORD) by padding. Padding bytes (not necessarily 0) must be appended to the end of the rows in order to bring up the length of the rows to a multiple of four bytes. When the pixel array is loaded into memory, each row must begin at a memory address that is a multiple of 4. This address/offset restriction is mandatory only for Pixel Arrays loaded in memory. For file storage purposes, only the size of each row must be a multiple of 4 bytes while the file offset can be arbitrary. A 24-bit bitmap with Width=1, would have 3 bytes of data per row (blue, green, red) and 1 byte of padding, while Width=2 would have 2 bytes of padding, Width=3 would have 3 bytes of padding, and Width=4 would not have any padding at all.
The wikipedia article on Data Structure Padding is also an interesting read that explains the reasons that paddings are generally used in computer science.
I presume this was design decision to align for better memory patterns while not wasting that much space (for 319px wide image you would waste 3 bytes or 0.25%)
Imagine you need to access some odd row directly. You could access first 4 pixels of n-th row by doing:
uint8_t *startRow = bmp + n * width * 3; //3 bytes per pixel
uint8_t r1 = startRow[0];
uint8_t g1 = startRow[1];
//... Repeat
uint8_t b4 = startRow[11];
Note that if n and width are odd (and bmp is even), startRow is going to be odd.
Now if you tried to do following speedup:
uint32_t *startRow = (uint32_t *) (bmp + n * width * 3);
uint32_t a = startRow[0]; //Loading register at a time is MUCH faster
uint32_t b = startRow[1]; //but only if address is aligned
uint32_t c = startRow[2]; //else code can hit bus errors!
uint8_t r1 = (a & 0xFF000000) >> 24;
uint8_t g1 = (a & 0x00FF0000) >> 16;
//... Repeat
uint8_t b4 = (c & 0x000000FF) >> 0;
You'd run into lots of problems. In best case scenario (that is intel cpu) your every load of a, b and c would need to be broken into two loads since startRow is not divisible by 4. In worst case scenario (eg. sun sparc) your program would crash with "bus error".
In newer designs it is common to force rows to be aligned to at least L1 cache line size (64 bytes on intel or 128 bytes on nvidia gpus).
Short version
Because the bmp file format specifies rows must perfectly fit in a 32bits "memory cells". Because pixels are 24bits, some combinations of pixels will not perfect sit in 32bit "cells". In this case, the cell is "padded up to" the full 32bits.
8bits per byte ∴
cell: 32bit = 4bytes ∴
pixel: 24bits = 3bytes
// If doesn't fit perfectly in 4 byte "cell"
if( read % 4 != 0 ) {
// find the difference between the "cell", and "the partial fit"
read = 4 - (read%4);
printf( "Padding: %d bytes\n", read );
// skip the difference
fread( pixel, read, 1, inFile );
}
Long version
In computing, a word is the natural unit of data used by a particular processor design. A word is a fixed-sized piece of data handled as a unit by the instruction set or the hardware of the processor
-wiki: Word_(computer_architecture)
Computer systems basically have a preferred "word length" (though not so important these days). A standard data unit allows all sorts of optimisations in the architecture of the computer system (think what shipping containers did for the shipping industry). There is a 32 bit standard called DWORD aka Double word (I guess) - and thats what typical bitmap images are optimised for.
So if you have 24bits per pixel, there will be various "literal pixels" row lengths that will not fit nicely into the 32bits. So in that case, pad it out.
Note: today, you are probably using a computer with a 64bit word size. Check your processor.
It depends on the format whether or not there is padding at the end of each row.
There really isn't much reason for it for 3 x 8 bit channel images since I/O is byte oriented anyway. For images with pixels packed into less than a byte (1 bit / pixel for example), padding is useful so that each row starts at a byte offset.