Can we say that our 'traditional' way of writing in binary
is Big Endian?
e.g., number 1 in binary:
0b00000001 // Let's assume its possible to write numbers like that in code and b means binary
Also when I write a constant 0b00000001 in my code, this will always refer to integer 1 regardless if machine is big endian or little endian right?
In this notation the LSB is always written as the last element from the right, and MSB is always written as the left most element right?
Yes, humans generally write numerals in big-endian order (meaning that the digits written first have the most significant value), and common programming languages that accept numerals interpret them in the same way.
Thus, the numeral “00000001” means one; it never means one hundred million (in decimal) or 128 (in binary) or the corresponding values in other bases.
Much of C semantics is written in terms of the value of a number. Once a numeral is converted to a value, the C standard describes how that value is added, multiplied, and even represented as bits (with some latitude regarding signed values). Generally, the standard does not specify how those bits are stored in memory, which is where endianness in machine representations comes into play. When the bits representing a value are grouped into bytes and those bytes are stored in memory, we may see those bytes written in different orders on different machines.
However, the C standard specifies a common way of interpreting numerals in source code, and that interpretation is always big-endian in the sense that the most significant digits appear first.
If you want to put it that way, then yes, we humans write numerals in Big-Endian order.
But I think you have a misunderstanding in terms of your target runnign with big or little endian.
In your actual C-Code, it does not matter which endianess your target machine uses. For example these lines will always display the same, no matter the endianess of your system:
uint32 x = 0x0102;
printf("Output: %x\n",x); // Output: 102
or to take your example:
uint32 y = 0b0001;
printf("Output: %d\n",y); // Output: 1
However the storage of the data in your memory differs between Little and Big Endian.
Big Endian:
Actual Value: 0x01020304
Memory Address: 0x00 0x01 0x02 0x03
Value: 0x01 0x02 0x03 0x04
Little Endian:
Actual Value: 0x01020304
Memory Address: 0x00 0x01 0x02 0x03
Value: 0x04 0x03 0x02 0x01
Both times the actualy value is 0x01020304 (and this is what you assign in your C-Code).
You only have to worry about it, if you do memory operations. If you have a 4-Byte (uint8) array, which represents a 32-Bit integer and you want to copy it into a uint32 variable you need to care.
uint8 arr[4] = {0x01, 0x02, 0x03, 0x04};
uint32 var;
memcpy(&var,arr,4);
printf("Output: %x\n",var);
// Big Endian: Output: 0x01020304
// Little Endian: Output: 0x04030201
Related
I have u16 field in a code that assigned value like 0x1f does this means 0x1f00? can any one please tell me this. thanks. so 1f in upper half 8 bits of 16 bit field.
like 1f00
or 0001111100000000
No, 0x1f is just another way of writing 31, so it does NOT mean 0x1f00, which is another way of writing 7936. (Tip: type "0x1f00 in decimal" into Google to verify this for yourself.)
It is correct to say that 0x1f is the same as 0x001f.
There is no standard data type named u16, but I suppose you might be talking about uint16_t from stdint.h. The C standard does not say exactly how a uint16_t is encoded as bits on your target machine, but most computers use a convention called "little endian", which means the number will be stored as two bytes, and the least-significant byte is first. So a uint16_t with a value of 0x1f (also known as 0x001f) would be encoded as two bytes, in this order: 0x1f 0x00.
You can experiment with this for yourself by making a struct with some uint16_t members in them, setting them to different values, and then casting a pointer to that struct to type uint8_t *. Then use the pointer to read the byte values at different locations in the struct, to see exactly how your struct is stored in memory.
I am a bit confused on how you would approach this problem:
Consider decimal number 1027. This value is stored as a 16-bit two's complement number into addresses 124 and 125 on a little endian machine which has an addressable cell size of one byte. What values (in hexadecimal) are in each of these addresses:
124:
125:
I know that a little endian machine orders it addresses from the least significant byte to the most significant byte. But besides from that I am unsure of how you would apply that concept and how you would order bytes into the addresses.
Here's some simple Python code to convert that integer to little-endian hexadecimal representation:
# convert the integer (1027) to hex using 2 bytes and little-endian byteorder
(1027).to_bytes(length=2, byteorder='little').hex()
This gives 0304. So, the first byte (03) is in address 124 and the second one (04) occupies the next address - 125.
"Little endian" and "big endian" relate to how the machine multiplexes bytes from memory into registers of the CPU.
With each byte it gets, it increments the address counter, but does it place these bytes from left-to-right or right-to-left into the register?
So the address that gets loaded into a machine register (or an integer), can be stored reverse in the memory. Even with modern CPUs with broad data busses, the concept remained and in some CPUs the bytes get swapped inside the CPU.
i am looking at the CoAP implementation in ContikiOS, particularly at the header parsing and i am having some trouble understanding two operations. The code is bellow.
coap_pkt->version = (COAP_HEADER_VERSION_MASK & coap_pkt->buffer[0]) >> COAP_HEADER_VERSION_POSITION;
coap_pkt->type = (COAP_HEADER_TYPE_MASK & coap_pkt->buffer[0]) >> COAP_HEADER_TYPE_POSITION;
The coap_pkt the structure that houses the packet and the parsed values.
the version is an uint8_t (unsigned char) type
The buffer houses the packet
The COAP_HEADER_VERSION_MASK value is 0xC0
COAP_HEADER_VERSION_POSITION is 6
the type value is an enum structure with 4 values
COAP_HEADER_TYPE_MASK value is 0x30
COAP_HEADER_TYPE_POSITION value is 4
Now according to CoAP RFC 7252 the both the version and the type occupy two bits and so the mask operations and used values make sense to me.
My question is why are the shifting values different in the two operations ?,does it have something to do with one using an unsigned char and another the enum structure?
Basically this shifts both of the values down to the least significant bits.
The version bits are at the most significant bits, let's call them bits 7 and 6. So, the four possible values with the mask applied are 0xC0, 0x80, 0x40 and 0x00. For later use, e.g. for version comparisons, changing the range to 3, 2, 1 and 0 might be more useful, so shifting by six bits moves the value to bits 1 and 0. (In fact the mask is unnecessary in this case because the shifting eliminates all but the most significant two bits.)
It's the same story with the type bits, but those are bits 5 and 4, giving you 0x30, 0x20, 0x10 and 0x00 after applying the mask. Shifting by four bits moves those bits to 1 and 0.
Number 4 represented as a 32-bit unsigned integer would be
on a big endian machine:
00000000 00000000 00000000 00000100 (most significant byte first)
on a small endian machine:
00000100 00000000 00000000 00000000 (most significant byte last)
As a 8-bit unsigned integer it is represented as
00000100 on both machines.
Now when casting 8-bit uint to a 32-bit I always thought that on a big endian machine that means sticking 24 zeros in front of the existing byte, and appending 24 zeros to the end if the machine is little endian. However, someone pointed out that in both cases zeros are prepended rather than appended. But wouldn't it mean that on a little endian 00000100 will become the most significant byte, which will result in a very large number? Please explain where I am wrong.
Zeroes are prepended if you consider the mathematical value (which just happens to also be the big-endian representation).
Casts in C always strive to preserve the value, not representation. That's how, for example, (int)1.25 results(*note below) in 1, as opposed to something which makes much less sense.
As discussed in the comments, the same holds for bit-shifts (and other bitwise operations, for that matter). 50 >> 1 == 25, regardless of endianness.
(* note: usually, depends rounding mode for float->integer conversion)
In short: Operators in C operate on the mathematical value, regardless of representation. One exception is when you cast a pointer to the value (as in (char*)&foo), since then it is essentially a different "view" to the same data.
Not sure if it answers your question, but will give it a try:
If you take a char variable and cast it to an int variable, then you get the exact same result on both architectures:
char c = 0x12;
int i = (int)c; // i == 0x12 on both architectures
If you take an int variable and cast it to a char variable, then you get the exact same result (possibly truncated) on both architectures:
int i = 0x12345678;
char c = (char)i; // c == 0x78 on both architectures
But if you take an int variable and read it using a char* pointer, then you get a different result on each architecture:
int i = 0x12345678;
char c = *(char*)&i; // c == 0x12 on BE architecture and 0x78 on LE architecture
The example above assumes that sizeof(int) == 4 (may be different on some compilers).
Loosely speaking, "Endianness" is the property of how processor sees the data stored in memory. This means that all the processors, when a particular data is brought to the CPU, sees it the same way.
For example:
int a = 0x01020304;
Irrespective of whether a little or big endian machine, would always have 04 as the least significant and 01 as the most significant byte, when stored in it's register.
The problem arises when this variable/data has to be stored in memory, which is "byte addressable". Should 01 (Most Significant Byte) go into the lowest memory address (Big Endian) or the highest memory address (Little Endian).
In your particular example, what you have shown is the representation, the way processor sees it - with LS/MS Byte.
So technically speaking, both little and big endian machines would have:
00000000 00000000 00000000 00000100
in its 32 bit wide register. Assuming of course what you have in memory is 32 bit wide integer representing 4. How this 4 is stored in/retrieved from memory is what endianness is all about.
I'm trying to read a binary file into a C# struct. The file was created from C and the following code creates 2 bytes out of the 50+ byte rows.
unsigned short nDayTimeBitStuffed = atoi( LPCTSTR( strInput) );
unsigned short nDayOfYear = (0x01FF & nDayTimeBitStuffed);
unsigned short nTimeOfDay = (0x01F & (nDayTimeBitStuffed >> 9) );
Binary values on the file are 00000001 and 00000100.
The expected values are 1 and 2, so I think some bit ordering/swapping is going on but not sure.
Any help would be greatly appreciated.
Thanks!
The answer is 'it depends' - most notably on the machine, and also on how the data is written to the file. Consider:
unsigned short x = 0x0102;
write(fd, &x, sizeof(x));
On some machines (Intel), the low-order byte (0x02) will be written before the high-order byte (0x01); on others (PPC, SPARC), the high-order byte will be written before the low-order one.
So, from a little-endian (Intel) machine, you'd see the bytes:
0x02 0x01
But from a big-endian (PPC) machine, you'd see the bytes:
0x01 0x02
Your bytes appear to be 0x01 and 0x04. Your calculation for 0x02 appears flawed.
The C code you show doesn't write anything. The value in nDayOfYear is the bottom 9 bits of the input value; the nTimeOfDay appears to be the next 5 bits (so 14 of the 16 bits are used).
For example, if the value in strInput is 12141 decimal, 0x2F6D, then the value in nDayOfYear would be 365 (0x16D) and the value in nTimeOfDay would be 23 (0x17).
It is a funny storage order; you can't simply compare the two values whereas if you packed the day of year in the more significant portion of the value and time into the less significant, then you could compare values as simple integers and get the correct comparison.
The expected file contents are very much related to the processor and compiler used to create the file, if it's binary.
I'm assuming a Windows machine here, which uses 2 bytes for a short and puts them in little endian order.
Your comments don't make much sense either. If it's two bytes then it should be using two chars, not shorts. The range of the first is going to be 1-365, so it definitely needs more than a single byte to represent. I'm going to assume you want the first 4 bytes, not the first 2.
This means that the first byte will be bits 0-7 of the DayOfYear, the second byte will be bits 8-15 of the DayOfYear, the third byte will be bits 0-7 of the TimeOfDay, and the fourth byte will be bits 8-15 of the TimeOfDay.