Recently I've learned some Windows socket programming to achieve some socket connection stuff.
Inside the code we use some functions like htonl(), htons() to convert our data from host byte order to network byte order, or so called big-endian, since in some machines data is stored as little-endian, like Intel's CPU as I know.
But what I confused is that doesn't the really important thing is the bit order instead of byte?
Since bit is the minimum unit that computers used, not byte.
Let's say we want to pass an u_short u=18 to another machine by a machine use little-endian.
In our machine, u's first byte, the least significant one is 2 and second byte, the most significant one is 1, let me express this by [2][1].
So we need to call htons(u) first to convert [2][1] into [1][2] and send it to network.
The remote machine will receive 2 first, but 2 is expressed by a sequence of bits which the machine actually receive are 0000-0010, how does the machine know this byte is 2? Won't it think 0000-0010 represents 64?(0100-0000) Do all machines store bytes in the same way?
"Bit Ordering" is specified by the protocols used to transfer data from the hardware (like your network card/port) to host memory.
Technically, this applies in many, many cases. Like, you might wonder the same question if a Harddrive on a machine that stores bits in one order 01234567 is transferred to a machine that stores bits in the opposite order 76543210, whether it'll read the data correctly or not.
But the simple answer is that it always reads correctly, because the protocols used to map the hard drives to the system bus specify the exact ordering of the bits as they are "presented" to host memory.
Network cards and networking hardware have a similar behavior: They have a standardized "bit ordering" they use in hardware, and as part of the hardware protocol, it "presents" those bits to the Host in whatever form the host expects them.
"Byte Ordering", of course, is a separate thing, and more difficult to deal with, because network hardware and storage hardware only recognize "streams of bytes" (Yes, I'm oversimplifying), and don't much care for what those bytes actually mean.
But in terms of "Bit Ordering", unless you're writing code "On the Metal", so to speak, you don't need to think about it.
Related
I am having a doubt about how to store IPv6 addresses on my system. On my system I accept a packet from the network run statistics on it and send it forward.
I have read the that common way to store it is in big-endian(network order) regardless of the endianness of the CPU. Although several colleagues of mine said they also familiar with this notation no one including google couldn't give a concrete explanation of why its the customary tradition.
The way I see it, the data will be accessed many times in the system so isn't it simpler to change it to host-order(on my case its little-endian) and use it on the entire system without having to run ntoh calls each time I want to run some arithmetic operation on it, and simply change it to network-order if I want to use this address on a packet I want to send back on the network?
The reason is as simple as: because RFC-1700 says so.
Following a three-decade dispute over which endianness is the single "correct one", Danny Cohen published On Holy Wars and a Plea of Peace in an attempt to shed light into the problem (which is deeper than just "byte order") and with the futile hope that the industry would agree on one consistent order.
The bottom line was that as long as you transmit a message as "one message", there is no such problem as order, but as soon you transmit sub-parts in the message (words, bytes, or bits) you need to decide for one order. It does not matter much which one you choose as long as you stick to your decision. The debate about which order is more correct than the other was as unproductive and silly as the debate on how to break an egg in "Gulliver's Travels", which Cohen referred to and took the name "endianness" from.
In 1994, the authors of RFC-1700 decided to end the debate, at least as far as the IP suite was concerned, by stating:
The convention in the documentation of Internet Protocols is to express numbers in decimal and to picture data in "big-endian" order. That is, fields are described left to right, with the most significant octet on the left and the least significant octet on the right.
Every RFC thereafter has followed (explicitly or silently) that convention, which of course includes IPv6.
Practically speaking, it doesn't matter what byte order IP addresses are. They could be words or bytes in little-endian or big-endian, or nibbles, or they could be counting Quatloos if the implementors deemed that practical.
For 99% of all people, 99% of the time, it makes zero difference because neither are you supposed to look at or understand the address, nor do you need to remember "magic values", nor do you (normally) need to modify an address.
You usually get an opaque block of memory that is an IP address and port from "somewhere" (say, getaddrinfo or recvfrom) and you use that blob as-is, e.g. with socket or sendto.
You do not normally need to perform "math" on the address, or anything similar. At best, you might want to compare two addresses for equality.
Yes, applications that need to perform more complicated things with addresses exist, but they are by far the minority of applications.
The way I see it, the data will be accessed many times in the system so isn't it simpler to change it to host-order(on my case its little-endian) and use it on the entire system without having to run ntoh calls each time I want to run some arithmetic operation on it, and simply change it to network-order if I want to use this address on a packet I want to send back on the network?
Well, you're right to wonder about this, but I believe that what you call customary tradition, is just a design choice when doing an application.
I might reverse your question: when you don't need to do any arithmetics on the addresses, why get into the trouble of reversing the order twice just for the sake of keeping it into the host's order?
If your application does a lot of maths on the IP address, it might be indeed smart to reverse it to little-endianness and reverse it before sending it back. But if you do not do any maths on it, then just keep it big-endian. And don't forget that the x86 is not the only CPU around, you also got other host architectures, like ARM or PPC, that were big-endian until they became bi-endian.
I read on internet that standard byte order for networks is big endian, also known as network byte order. Before transferring data on network, data is first converted to network byte order (big endian).
But can any one please let me know who will take care of this conversion.
Whether the code developer do really worry about this endianness? If yes, can you please let me know the examples where we need to take care (in case of C, C++).
The first place where the network vs native byte order matters is in creating sockets and specifying the IP address and port number. Those must be in the correct order or you will not end up talking to the correct computer, or you'll end up talking to the incorrect port on the correct computer if you mapped the IP address but not the port number.
The onus is on the programmer to get the addresses in the correct order. There are functions like htonl() that convert from host (h) to network (n) order; l indicates 'long' meaning '4 bytes'; s indicates 'short' meaning '2 bytes' (the names date from an era before 64-bit systems).
The other time it matters is if you are transferring binary data between two computers, either via a network connection correctly set up over a socket, or via a file. With single-byte code sets (SBCS), or UTF-8, you don't have problems with textual data. With multi-byte code sets (MBCS), or UTF-16LE vs UTF-16BE, or UTF-32, you have to worry about the byte order within characters, but the characters will appear one after the other. If you ship a 32-bit integer as 32-bits of data, the receiving end needs to know whether the first byte is the MSB (most significant byte — for big-endian) or the LSB (least significant byte — for little-endian) of the 32-bit quantity. Similarly with 16-bit integers, or 64-bit integers. With floating point, you could run into the additional problem that different computers could use different formats for the floating point, independently of the endianness issue. This is less of a problem than it used to be thanks to IEEE 744.
Note that IBM mainframes use EBCDIC instead of ASCII or ISO 8859-x character sets (at least by default), and the floating point format is not IEEE 744 (pre-dating that standard by a decade or more). These issues, therefore, are crucial to deal with when communicating with the mainframe. The programs at the two ends have to agree with how each end will understand the other. Some protocols define a byte order (e.g. network byte order); others define 'sender makes right' or 'receiver makes right' or 'client makes right' or 'server makes right', placing the conversion workload on different parts of the system.
One advantage of text protocols (especially those using an SBCS) is that they evade the problems of endianness — at the cost of converting text to value and back, but computation is cheap compared to even gigabit networking speeds.
In C and C++, you will have to worry about endianness in low level network code. Typically the serialization and deserialization code will call a function or macro that adjusts the endianness - reversing it on little endian machines, doing nothing on big endian machines - when working with multibyte data types.
Just send stuff in the correct order that the receiver can understand,
i.e. use http://www.manpagez.com/man/3/ntohl/ and their ilk.
We are all fans of portable C/C++ programs.
We know that sizeof(char) or sizeof(unsigned char) is always 1 "byte". But that 1 "byte" doesn't mean a byte with 8 bits. It just means a "machine byte", and the number of bits in it can differ from machine to machine. See this question.
Suppose you write out the ASCII letter 'A' into a file foo.txt. On any normal machine these days, which has a 8-bit machine byte, these bits would get written out:
01000001
But if you were to run the same code on a machine with a 9-bit machine byte, I suppose these bits would get written out:
001000001
More to the point, the latter machine could write out these 9 bits as one machine byte:
100000000
But if we were to read this data on the former machine, we wouldn't be able to do it properly, since there isn't enough room. Somehow, we would have to first read one machine byte (8 bits), and then somehow transform the final 1 bit into 8 bits (a machine byte).
How can programmers properly reconcile these things?
The reason I ask is that I have a program that writes and reads files, and I want to make sure that it doesn't break 5, 10, 50 years from now.
How can programmers properly reconcile these things?
By doing nothing. You've presented a filesystem problem.
Imagine that dreadful day when the first of many 9-bit machines is booted up, ready to recompile your code and process that ASCII letter A that you wrote to a file last year.
To ensure that a C/C++ compiler can reasonably exist for this machine, this new computer's OS follows the same standards that C and C++ assume, where files have a size measured in bytes.
...There's already a little problem with your 8-bit source code. There's only about a 1-in-9 chance each source file is a size that can even exist on this system.
Or maybe not. As is often the case for me, Johannes Schaub - litb has pre-emptively cited the standard regarding valid formats for C++ source code.
Physical source file characters are mapped, in an
implementation-defined manner, to the basic source character set
(introducing new-line characters for end-of-line indicators) if
necessary. Trigraph sequences (2.3) are replaced by corresponding
single-character internal representations. Any source file character
not in the basic source character set (2.2) is replaced by the
universal-character-name that des- ignates that character. (An
implementation may use any internal encoding, so long as an actual
extended character encountered in the source file, and the same
extended character expressed in the source file as a
universal-character-name (i.e. using the \uXXXX notation), are handled
equivalently.)
"In an implementation-defined manner." That's good news...as long as some method exists to convert your source code to any 1:1 format that can be represented on this machine, you can compile it and run your program.
So here's where your real problem lies. If the creators of this computer were kind enough to provide a utility to bit-extend 8-bit ASCII files so they may be actually stored on this new machine, there's already no problem with the ASCII letter A you wrote long ago. And if there is no such utility, then your program already needs maintenance and there's nothing you could have done to prevent it.
Edit: The shorter answer (addressing comments that have since been deleted)
The question asks how to deal with a specific 9-bit computer...
With hardware that has no backwards-compatible 8-bit instructions
With an operating system that doesn't use "8-bit files".
With a C/C++ compiler that breaks how C/C++ programs have historically written text files.
Damian Conway has an often-repeated quote comparing C++ to C:
"C++ tries to guard against Murphy, not Machiavelli."
He was describing other software engineers, not hardware engineers, but the intention is still sound because the reasoning is the same.
Both C and C++ are standardized in a way that requires you to presume that other engineers want to play nice. Your Machiavellian computer is not a threat to your program because it's a threat to C/C++ entirely.
Returning to your question:
How can programmers properly reconcile these things?
You really have two options.
Accept that the computer you describe would not be appropriate in the world of C/C++
Accept that C/C++ would not be appropriate for a program that might run on the computer you describe
Only way to be sure is to store data in text files, numbers as strings of number characters, not some amount of bits. XML using UTF-8 and base 10 should be pretty good overall choice for portability and readability, as it is well defined. If you want to be paranoid, keep the XML simple enough, so that in a pinch it can be easily parsed with simple custom parser, in case a real XML parser is not readily available for your hypothetical computer.
When parsing numbers, and it is bigger than what fits in your numeric data type, well, that's an error situation you need to handle as you see fit in the context. Or use a "big int" library, which can then handle arbitrarily large numbers (with an order of magnitude performance hit compared to "native" numeric data types, of course).
If you need to store bit fields, then store bit fields, that is number of bits and then bit values in whatever format.
If you have a specific numeric range, then store the range, so you can explicitly check if they fit in available numeric data types.
Byte is pretty fundamental data unit, so you can not really transfer binary data between storages with different amount of bits, you have to convert, and to convert you need to know how the data is formatted, otherwise you simply can not convert multi-byte values correctly.
Adding actual answer:
In you C code, do not handle byte buffers, except in isolated functions which you will then modify as appropriate for CPU architecture. For example .JPEG handling functions would take either a struct wrapping the image data in unspecified way, or a file name to read the image from, but never a raw char* to byte buffer.
Wrap strings in a container which does not assume encoding (presumably it will use UTF-8 or UTF-16 on 8-bit byte machine, possibly currently non-standard UTF-9 or UTF-18 on 9-bit byte machine, etc).
Wrap all reads from external sources (network, disk files, etc) into functions which return native data.
Create code where no integer overflows happen, and do not rely on overflow behavior in any algorithm.
Define all-ones bitmasks using ~0 (instead of 0xFFFFFFFF or something)
Prefer IEEE floating point numbers for most numeric storage, where integer is not required, as those are independent of CPU architecture.
Do not store persistent data in binary files, which you may have to convert. Instead use XML in UTF-8 (which can be converted to UTF-X without breaking anything, for native handling), and store numbers as text in the XML.
Same as with different byte orders, except much more so, only way to be sure is to port your program to actual machine with different number of bits, and run comprehensive tests. If this is really important, then you may have to first implement such a virtual machine, and port C-compiler and needed libraries for it, if you can't find one otherwise. Even careful (=expensive) code review will only take you part of the way.
if you're planning to write programs for Quantum Computers(which will be available in the near future for us to buy), then start learning Quantum Physics and take a class on programming them.
Unless you're planning for a boolean computer logic in the near future, then.. my question is how will you make it sure that the filesystem available today will not be the same tomorrow? or how a file stored with 8 bit binary will remain portable in the filesystems of tomorrow?
If you want to keep your programs running through generations, my suggestion is create your own computing machine, with your own filesystem and your own operating system, and change the interface as the needs of tomorrow change.
My problem is, the computer system I programmed a few years ago doesn't exist(Motorola 68000) anymore for normal public, and the program heavily relied on the machine's byte order and assembly language. Not portable anymore :-(
If you're talking about writing and reading binary data, don't bother. There is no portability guarantee today, other than that data you write from your program can be read by the same program compiled with the same compiler (including command-line settings). If you're talking about writing and reading textual data, don't worry. It works.
First: The original practical goal of portability is to reduce work; therefore if portability requires more effort than non-portability to achieve the same end result, then writing portable code in such case is no longer advantageous. Do not target 'portability' simply out of principle. In your case, a non-portable version with well-documented notes regarding the disk format is a more efficient means of future-proofing. Trying to write code that somehow caters to any possible generic underlying storage format will probably render your code nearly incomprehensible, or so annoying to maintain that it will fall out of favor for that reason (no need to worry about future-proofing if no one wants to use it anyway 20 yrs from now).
Second: I don't think you have to worry about this, because the only realistic solution to running 8-bit programs on a 9-bit machine (or similar) is via Virtual Machines.
It is extremely likely that anyone in the near or distant future using some 9+ bit machine will be able to start up a legacy x86/arm virtual machine and run your program that way. Hardware 25-50 years from now should have no problem what-so-ever of running entire virtual machines just for the sake of executing a single program; and that program will probably still load, execute, and shutdown faster than it does today on current native 8-bit hardware. (some cloud services today in fact, already trend toward starting entire VMs just to service individual tasks)
I strongly suspect this is the only means by which any 8-bit program would be run on 9/other-bit machines, due to the points made in other answers regarding the fundamental challenges inherent to simply loading and parsing 8-bit source code or 8-bit binary executables.
It may not be remotely resembling "efficient" but it would work. This also assumes, of course, that the VM will have some mechanism by which 8-bit text files can be imported and exported from the virtual disk onto the host disk.
As you can see, though, this is a huge problem that extends well beyond your source code. The bottom line is that, most likely, it will be much cheaper and easier to update/modify or even re-implement-from-scratch your program on the new hardware, rather than to bother trying to account for such obscure portability issues up-front. The act of accounting for it almost certainly requires more effort than just converting the disk formats.
8-bit bytes will remain until end of time, so don't sweat it. There will be new types, but this basic type will never ever change.
I think the likelihood of non-8-bit bytes in future computers is low. It would require rewriting so much, and for so little benefit. But if it happens...
You'll save yourself a lot of trouble by doing all calculations in native data types and just rewriting inputs. I'm picturing something like:
template<int OUTPUTBITS, typename CALLABLE>
class converter {
converter(int inputbits, CALLABLE datasource);
smallestTypeWithAtLeast<OUTPUTBITS> get();
};
Note that this can be written in the future when such a machine exists, so you need do nothing now. Or if you're really paranoid, make sure get just calls datasource when OUTPUTBUTS==inputbits.
Kind of late but I can't resist this one. Predicting the future is tough. Predicting the future of computers can be more hazardous to your code than premature optimization.
Short Answer
While I end this post with how 9-bit systems handled portability with 8-bit bytes this experience also makes me believe 9-bit byte systems will never arise again in general purpose computers.
My expectation is that future portability issues will be with hardware having a minimum of 16 or 32 bit access making CHAR_BIT at least 16.
Careful design here may help with any unexpected 9-bit bytes.
QUESTION to /. readers: is anyone out there aware of general purpose CPUs in production today using 9-bit bytes or one's complement arithmetic? I can see where embedded controllers may exist, but not much else.
Long Answer
Back in the 1990s's the globalization of computers and Unicode made me expect UTF-16, or larger, to drive an expansion of bits-per-character: CHAR_BIT in C. But as legacy outlives everything I also expect 8-bit bytes to remain an industry standard to survive at least as long as computers use binary.
BYTE_BIT: bits-per-byte (popular, but not a standard I know of)
BYTE_CHAR: bytes-per-character
The C standard does not address a char consuming multiple bytes. It allows for it, but does not address it.
3.6 byte: (final draft C11 standard ISO/IEC 9899:201x)
addressable unit of data storage large enough to hold any member of the basic character set of the execution environment.
NOTE 1: It is possible to express the address of each individual byte of an object uniquely.
NOTE 2: A byte is composed of a contiguous sequence of bits, the number of which is implementation-defined. The least significant bit is called the low-order bit; the most significant bit is called the high-order bit.
Until the C standard defines how to handle BYTE_CHAR values greater than one, and I'm not talking about “wide characters”, this the primary factor portable code must address and not larger bytes. Existing environments where CHAR_BIT is 16 or 32 are what to study. ARM processors are one example. I see two basic modes for reading external byte streams developers need to choose from:
Unpacked: one BYTE_BIT character into a local character. Beware of sign extensions.
Packed: read BYTE_CHAR bytes into a local character.
Portable programs may need an API layer that addresses the byte issue. To create on the fly and idea I reserve the right to attack in the future:
#define BYTE_BIT 8 // bits-per-byte
#define BYTE_CHAR (CHAR_BIT/BYTE_BIT) //bytes-per-char
size_t byread(void *ptr,
size_t size, // number of BYTE_BIT bytes
int packing, // bytes to read per char
// (negative for sign extension)
FILE *stream);
size_t bywrite(void *ptr,
size_t size,
int packing,
FILE *stream);
size number BYTE_BIT bytes to transfer.
packing bytes to transfer per char character. While typically 1 or BYTE_CHAR it could indicate BYTE_CHAR of the external system, which can be smaller or larger than the current system.
Never forget endianness clashes.
Good Riddance To 9-Bit Systems:
My prior experience with writing programs for 9-bit environments lead me to believe we will not see such again, unless you happen to need a program to run on a real old legacy system somewhere. Likely in a 9-bit VM on a 32/64-bit system. Since year 2000 I sometimes make a quick search for, but have not seen, references to current current descendants of the old 9-bit systems.
Any, highly unexpected in my view, future general purpose 9-bit computers would likely either have an 8-bit mode, or 8-bit VM (#jstine), to run programs under. The only exception would be special purpose built embedded processors, which general purpose code would not likely to run on anyway.
In days of yore one 9-bit machine was the PDP/15. A decade of wrestling with a clone of this beast make me never expect to see 9-bit systems arise again. My top picks on why follow:
The extra data bit came from robbing the parity bit in core memory. Old 8-bit core carried a hidden parity bit with it. Every manufacturer did it. Once core got reliable enough some system designers switched the already existing parity to a data bit in a quick ploy to gain a little more numeric power and memory addresses during times of weak, non MMU, machines. Current memory technology does not have such parity bits, machines are not so weak, and 64-bit memory is so big. All of which should make the design changes less cost effective then the changes were back then.
Transferring data between 8-bit and 9-bit architectures, including off-the-shelf local I/O devices, and not just other systems, was a continuous pain. Different controllers on the same system used incompatible techniques:
Use the low order 16-bits of 18 bit words.
Use the low-order 8 bits of 9-bit bytes where the extra high-order bit might be set to the parity from bytes read from parity sensitive devices.
Combine the low-order 6 bits of three 8-bit bytes to make 18 bit binary words.
Some controllers allowed selecting between 18-bit and 16-bit data transfers at run time. What future hardware, and supporting system calls, your programs would find just can't be predicted in advance.
Connecting to the 8-bit Internet will be horrid enough by itself to kill any 9-bit dreams someone has. They got away with it back then as machines were less interconnected in those times.
Having something other than an even multiple of 2 bits in byte-addressed storage brings up all sorts of troubles. Example: if you want an array of thousands of bits in 8-bit bytes you can unsigned char bits[1024] = { 0 }; bits[n>>3] |= 1 << (n&7);. To fully pack 9-bits you must do actual divides, which brings horrid performance penalties. This also applies to bytes-per-word.
Any code not actually tested on 9-bit byte hardware may well fail on it's first actual venture into the land of unexpected 9-bit bytes, unless the code is so simple that refactoring it in the future for 9-bits is only a minor issue. The prior byread()/bywrite() may help here but it would likely need an additional CHAR_BIT mode setting to set the transfer mode, returning how the current controller arranges the requested bytes.
To be complete anyone who wants to worry about 9-bit bytes for the educational experience may need to also worry about one's complement systems coming back; something else that seems to have died a well deserved death (two zeros: +0 and -0, is a source of ongoing nightmares... trust me). Back then 9-bit systems often seemed to be paired with one's complement operations.
In a programming language, a byte is always 8-bits. So, if a byte representation has 9-bits on some machine, for whatever reason, its up to the C compiler to reconcile that. As long as you write text using char, - say, if you write/read 'A' to a file, you would be writing/reading only 8-bits to the file. So, you should not have any problem.
I have come to believe that the optimal size for a boolean variable is the natural width of the data, ie in C/C++ it is int. So for modern processors this is normally 32 bits. At the machine level declaring it as a byte for example requires a 32 bit fetch and then a mask.
However I have seen that a BOOL in iOS is 8 bits. I had assumed that people who used bytes were using left-over ideas from 8 bit processors.
I realise this question depends on the use and for most of the time the language defined boolean is the best bet, but there are times when you need to define your own, such as when you are converting code arriving from an external source or you want to write cross platform code.
It is also significant that if a boolean value is going to be packed into a serial stream, for sending over a serial line such as ethernet or storing it may be optimal to pack the boolean in fewer bits. But I feel that it is likely that it is optimal to pack and unpack from a processor optimal size.
So my question is am I correct in thinking that the optimal size for a boolean on a 32bit processor is 32 bits and if so why does iOS use 8 bits.
Yup you are right it depends. The big advantage of using an 8-bit is that you can pack more into a struct nicely.
Of course you'd be best off using flags in such a case.
The big issue, though, is that with a C/C++ "bool" you don't necessarily know how big it is. This means that you can't make assumptions about a struct (such as binary writing to disk) without the possibility of it breaking on another platform. In such a case using a known sized variable can be very useful and you may as well use as little space as possible if you are going to dump the structure to disk.
The notion of an 8-bit quantity involving a 32-bit fetch followed by hardware masking is mostly obsolete. In reality, a fetch from memory (on a modern processor) will normally be one L2 cache line (typically around 64-128 bytes). That being the case, essentially every size of item you deal with involves fetching a big chunk of data, and then using only some subset of what you fetched (but, assuming your data is more or less contiguous, probably using more of that data subsequently).
C++ attempts (not necessarily successfully) to optimize this a bit for you. An individual bool can be anywhere from one byte on up, though on most typical implementation, it's either one byte or four bytes. The (much reviled) std::vector<bool> uses some tricks to give a (sort of) vector-like interface, but still store each bool in one bit. In the process it loses the ability to be treated as a generic sequence container -- but when you're storing a lot of bools, and can live with the restrictions of using it in an array-like manner, it can actually be a lot more useful than many people believe.
When/if you want to retain normal container semantics and don't mind the extra storage space to keep them their native size, you can use another container (e.g., std::deque<bool>) instead. Especially if you only need to store a small collection of bools, this can often be a superior alternative.
It is architecture dependent, but on many 32 bit architectures 8 bit addressing is no less efficient than 32 bit; the "fetching and masking" as such is performed in hardware logic.
The optimal size in terms of storage space is of course 1 bit. You might for example use bit-fields or bit masking to pack multiple booleans in a single word. Some architectures such as 8051 have bit addressable memory. The more modern ARM Cortex-M architecture employs a technique called bit-banding that allows memory and hardware registers to be bit addressable
Are there many computers which use Big Endian? I just tested on 5 different computers, each purchased in different years, and different models. Each use Little Endian. Is Big Endian still used now days or was it for older processors such as the Motorola 6800?
Edit:
Thank you TreyA, intel.com/design/intarch/papers/endian.pdf is a very nice and handy article. It covers every answers bellow, and also expands upon them.
There's many processors in use today that is big endian, or allows the option to switch endian mode between big and little endian, (e.g. SPARC, PowerPC, ARM, Itanium..).
It depends on what you mean by "care about endian". You usually don't need to care that much specifically about endianess if you just program to the data you need. Endian matters when you need to communicate to the outside world, such as read/write a file, or send data over a network and you do that by reading/writing integers larger than 1 byte directly to/from memory.
When you do need to deal with external data, you need to know its format. Part of its format is to e.g. know how an integer is encoded in that data. If the format specifies that the first byte of an 4 byte integer is the most significant byte of said integer, you read that byte and place it at the most significant byte of the integer in your program, and you would be able to accomplish that fine
with code that runs on both little and big endian machines.
So it's not so much specifically about the processor endianess, but the data you need to deal with. That data might have integers stored in either "endian", you need to know which, and various data formats will use various endianess depending on some specification, or depending on the whim of the guy that came up with the format.
Big endian is still by far the most used, in terms of different architectures. In fact, outside of the Intel and the old DEC computers, I don't know of a small endian: Sparc, Power PC (IBM Unix machines), HP's Unix platforms, IBM mainframes, etc. are all big endian. But does it matter? About the only time I've had to consider endianness was when implementing some low level system routines, like modf. Otherwise, int is an integer value in a certain range, and that's it.
The following common platforms use big-endian encoding:
Java
Network data in TCP/UDP packets (maybe even on the IP level, but I'm not sure about that)
The x86/x64 CPUs are little-endian. If you are going to interface with binary data between the two, you should definitely be aware of this.
This qualifies more as a comment than an answer, but I can't comment and I think it's such a great article to read, that I think it worthwhile.
This is a classic on endianness by Danny Cohen, dating from 1980:
ON HOLY WARS AND A PLEA FOR PEACE
There is not enough context to the question. In general, you should simply be aware of it at all times, but you do not need to stress over it in everyday coding. If you plan on messing with the internal bytes of your integers, start worrying about endianness. If you plan on doing standard math on your integers, don't worry about it.
The two big places where endianness pops up is in networking (big endian standard) and binary records (have to research whether integers are stored big endian or little endian).