An external group provides me with a file written on a Big Endian machine, and they also provide a C++ parser for the file format.
I only can run the parser on a little endian machine - is there any way to read the file using their parser without add a swapbytes() call after each read?
Back in the early Iron Age, the Ancients encountered this issue when they tried to network primitive PDP-11 minicomputers with other primitive computers. The PDP-11 was the first little-Endian computer, while most others at the time were big-Endian.
To solve the problem, once and for all, they developed the network byte order concept (always big-Endia), and the corresponding network byte order macros ntohs(), ntohl(), htons(), and htonl(). Code written with those macros will always "get the right answer".
Lean on your external supplier to use the macros in their code, and the file they supply you will always be big-Endian, even if they switch to a little-Endian machine. Rewrite the parser they gave you to use the macros, and you will always be able to read their file, even if you switch to a big-Endian machine.
A truly prodigious amount of programmer time has been wasted on this particular problem. There are days when I think a good argument could be made for hanging the PDP-11 designer who made the little-Endian feature decision.
Try persuading the parser team to include the following code:
int getInt(char* bytes, int num)
{
int ret;
assert(num == 4);
ret = bytes[0] << 24;
ret |= bytes[1] << 16;
ret |= bytes[2] << 8;
ret |= bytes[3];
return ret;
}
it might be more time consuming than a general int i = *(reinterpret_cast<*int>(&myCharArray)); but will always get the endianness right on both big and small endian systems.
In general, there's no "easy" solution to this. You will have to modify the parser to swap the bytes of each and every integer read from the file.
It depends upon what you are doing with the data. If you are going to print the data out, you need to swap the bytes on all the numbers. If you are looking through the file for one or more values, it may be faster to byte swap your comparison value.
In general, Greg is correct, you'll have to do it the hard way.
the best approach is to just define the endianess in the file format, and not say it's machine dependent.
the writer will have to write the bytes in the correct order regardless of the CPU it's running on, and the reader will have to do the same.
You could write a parser that wraps their parser and reverses the bytes, if you don't want to modify their parser.
Be conscious of the types of data being read in. A 4-byte int or float would need endian correction. A 4-byte ASCII string would not.
In general, no.
If the read/write calls are not type aware (which, for example fread and fwrite are not) then they can't tell the difference between writing endian sensitive data and endian insensitive data.
Depending on how the parser is structured you may be able to avoid some suffering, if the I/O functions they use are aware of the types being read/written then you could modify those routines apply the correct endian conversions.
If you do have to modify all the read/write calls then creating just such a routine would be a sensible course of action.
Your question somehow conatins the answer: No!
I only can run the parser on a little endian machine - is there any way to read the file using their parser without add a swapbytes() call after each read?
If you read (and want to interpret) big endian data on a little endian machine, you must somehow and somewhere convert the data. You might do this after each read or after the whole file has been read (if the data read does not contain any information on how to read further data) - but there is no way in omitting the conversion.
Related
I have 3 vector, each with exactly 256^3 ~ 16 million elements that i want to store in a file and read as fast as possible. I only care about reading performance, and the representation of the data in memory can be any.
I have taken a look at some serialization techniques as well as writing/ reading plain numbers to/ from a file with ofstream, however i wonder if there is a more direct and faster approach.
(i am pretty new to c++ and its concepts)
Assuming both systems, windows and android, are little endian, which is common in ARM and x86/x64 CPUs, you can do the following.
First: Determine the type with a sepcific size, so either double, with 64-bit, float with 32-bit, or uint64/32/16 or int64/32/16. Do NOT use stuff like int or long to determine your data type.
Second: Use the following method to write binary data:
std::vector<uint64_t> myVec;
std::ofstream f("outputFile.bin", std::ios::binary);
f.write(reinterpret_cast<char*>(myVec.data()), myVec.size()*sizeof(uint64_t));
f.close();
In this, you're take the raw data and writing its binary format to a file.
Now on other machine, make sure the data type you use has the same datatype size and same endianness. If both are the same, you can do this:
std::vector<uint64_t> myVec(sizeOfTheData);
std::ifstream f("outputFile.bin", std::ios::binary);
f.read(reinterpret_cast<char*>(&myVec.front()), myVec.size()*sizeof(uint64_t));
f.close();
Notice that you have to know the size of the data before reading it.
Note: This code is off my head. I haven't tested it, but it should work.
Now if the target system doesn't have the same endianness, you have to read the data in batches, flip the endianness, then put it in your vector. How to flip endianness was extensively discussed here.
To determine the endianness of your system, this was discussed here.
The penalty on performance will be proportional to how different these systems are. If they're both the same endianness and you choose the same data type and size, you're good and you have optimum performance. Otherwise, you'll have some penalty depending on how many conversion you have to do. This is the fastest that you can ever get.
Note from comments: If you're transferring doubles or floats, make sure both systems use IEEE 754 standard. It's very common to use these, way more than endianness, but just to be sure.
Now if these solutions don't fit you, then you have to use a proper serialization library to standardize the format for you. There are libraries that can do that, such as protobuf.
I know that this might be a silly question, but I am a newbie C++ developer and I need some clarifications about the endianness.
I have to implement a communication interface that relies on SCTP protocol in order to communicate between two different machines (one ARM based, and the other Intel based).
The aim is to:
encode messages into a stream of bytes to be sent on the socket (I used a vector of uint8_t, and positioned each byte of the different fields -taking care of splitting uint16/32/64 to single bytes- following big-endian convention)
send the bytestream via socket to the receiver (using stcp)
retrieve the stream and parse it in order to fill the message object with the correct elements (represented by header + TV information elements)
I am confused on where I could have problem with the endianness of the underlying architecture of the 2 machines in where the interface will be used.
I think that taking care of splitting objects into single bytes and positioning them using big-endian can preclude that, at the arrival, the stream is represented differently, right? or am I missing something?
Also, I am in doubt about the role of C++ representation of multiple-byte variables, for example:
uint16_t var=0x0123;
//low byte 0x23
uint8_t low = (uint8_t)var;
//hi byte 0x01
uint8_t hi = (uint8_t)(var >> 8);
This piece of code is endianness dependent or not? i.e. if I work on a big-endian machine I suppose that the above code is ok, but if it is little-endian, will I pick up the bytes in different order?
I've searched already for such questions but no one gave me a clear reply, so I have still doubts on this.
Thank you all in advance guys, have a nice day!
This piece of code is endianness dependent or not?
No the code doesn't depend on endianess of the target machine. Bitwise operations work the same way as e.g. mathematical operators do.
They are independent of the internal representation of the numbers.
Though if you're exchanging data over the wire, you need to have a defined byte order known at both sides. Usually that's network byte ordering (i.e. big endian).
The functions of the htonx() ntohx() family will help you do en-/decode the (multibyte) numbers correctly and transparently.
The code you presented is endian-independent, and likely the correct approach for your use case.
What won't work, and is not portable, is code that depends on the memory layout of objects:
// Don't do this!
uint16_t var=0x0123;
auto p = reinterpret_cast<char*>(&var);
uint8_t hi = p[0]; // 0x01 or 0x23 (probably!)
uint8_t lo = p[1]; // 0x23 or 0x01 (probably!)
(I've written probably in the comments to show that these are the likely real-world values, rather than anything specified by Standard C++)
Recently I was (again) reading about 'endian'ness. I know how to identify the endianness of host, as there are lots of post on SO, and also I have seen this, which I think is pretty good resource.
However, one thing I like to know is to how to detect the endianness of input binary file. For example, I am reading a binary file (using C++) like following:
ifstream mydata("mydata.raw", ios::binary);
short value;
char buf[sizeof(short)];
int dataCount = 0;
short myDataMat[DATA_DIMENSION][DATA_DIMENSION];
while (mydata.read(reinterpret_cast<char*>(&buf), sizeof(buf)))
{
memcpy(&value, buf, sizeof(value));
myDataMat[dataCount / DATA_DIMENSION][dataCount%DATA_DIMENSION] = value;
dataCount++;
}
I like to know how I can detect the endianness in the mydata.raw, and whether endianness affects this program anyway.
Additional Information:
I am only manipulating the data in myDataMat using mathematical operations, and no pointer operation or bitwise operation is done on the data).
My machine (host) is little endian.
It is impossible to "detect" the endianity of data in general. Just like it is impossible to detect whether the data is an array of 4 byte integers, or twice that many 2 byte integers. Without any knowledge about the representation, raw data is just a mass of meaningless bits.
However, with some extra knowledge about the data representation, it become possible. Some examples:
Most file formats mandate particular endianity, in which case this is never a problem.
Unicode text files may optionally start with a byte order mark. Same idea can be implemented by other data representations.
Some file formats contain a checksum. You can guess one endianity, and if the checksum does not match, try again with another endianity. It will be unlikely that the checksum matches with wrong interpretation of the data.
Sometimes you can make guesses based on the data. Is the temperature outside 33'554'432 degrees, or maybe 2? You can pick the endianity that represents sane data. Of course, this type of guesswork fails miserably, when the aliens invade and start melting our planet.
You can't tell.
The endianness transformation is essentially an operator E(x) on a number x such that x = E(E(x)). So you don't know "which way round" the x elements are in your file.
We have developed win32 application for x86 and x64 platform. We want to use the same application on ARM platform. Endianness will vary for ARM platform i.e. ARM platform uses Big endian format in general. So we want to handle this in our application for our device.
For e.g. // In x86/x64, int nIntVal = 0x12345678
In ARM, int nIntVal = 0x78563412
How values will be stored for the following data types in ARM?
double
char array i.e. char chBuffer[256]
int64
Please clarify this.
Regards,
Raphel
Endianess only matters for register <-> memory operations.
In a register there is no endianess. If you put
int nIntVal = 0x12345678
in your code it will have the same effect on any endianess machine.
all IEEE formats (float, double) are identical in all architectures, so this does not matter.
You only have to care about endianess in two cases:
a) You write integers to files that have to be transferable between the two architectures.
Solution: Use the hton*, ntoh* family of converters, use a non-binary file format (e.g. XML) or a standardised file format (e.g. SQLite).
b) You cast integer pointers.
int a = 0x1875824715;
char b = a;
char c = *(char *)&a;
if (b == c) {
// You are working on Little endian
}
The latter code by the way is a handy way of testing your endianess at runtime.
Arrays and the likes if you use write, fwrite falimies of calls to transfer them you will have no problems unless they contain integers: then look above.
int64_t: look above. Only care if you have to store them binary in files or cast pointers.
(Sergey L., above, says, taht you mostly don't have to care for the byte order. He is right, with at least 1 exception: I assumed you want to convert binary data from one platform to the other ...)
http://en.wikipedia.org/wiki/Endianness has a good overview.
In short:
Little endian means, the least significant byte is stored first (at the lowest address)
Big endian means the most significant byte is stored first
The order in which array elements are stored is not affected (but the byte order in array elements, of course)
So
char array is unchanged
int64 - byte order is reversed compared to x86
With regard to the floating point format, consider http://en.wikipedia.org/wiki/Endianness#Floating-point_and_endianness. Generally it seems to obey the same rules of endianness as the integer format, but there is are exceptions for older ARM platforms. (I've no first hand experience of that).
Generally I'd suggest, test your conversion of primitive types by controlled experiments first.
Also consider, that compilers might use different padding in structs (a topic you haven't addressed yet).
Hope this helps.
In 98% cases you don't need to care about endianness. Unless you need to transfer some data between systems of different endiannness, or read/write some endian-sensitive file format, you should not bother with it. And even in those cases, you can write your code to perform properly when compiled under any endianness.
From Rob Pike's "The byte order fallacy" post:
Let's say your data stream has a little-endian-encoded 32-bit integer.
Here's how to extract it (assuming unsigned bytes):
i = (data[0]<<0) | (data[1]<<8) | (data[2]<<16) | (data[3]<<24);
If it's big-endian, here's how to extract it:
i = (data[3]<<0) | (data[2]<<8) | (data[1]<<16) | (data[0]<<24);
Both these snippets work on any machine, independent of the machine's
byte order, independent of alignment issues, independent of just about
anything. They are totally portable, given unsigned bytes and 32-bit
integers.
The arm is little endian, it has two big endian variants depending on architecture, but it is better to just run native little endian, the tools and volumes of code out there are more fully tested in little endian mode.
Endianness is just one factor in system engineering if you do your system engineering it all works out, no fears, no worries. Define your interfaces and code to that design. Assuming for example that one processors endianness automatically results in having to byteswap is a bad assumption and will bite you eventually. You will end up having to swap an even number of times to undo other bad assumptions that cause a swap (ideally swapping zero times of course rather than 2 or 4 or 6, etc times). If you have any endian concerns at all when writing code you should write it endian independent.
Since some ARMs have BE32 (word invariant) and the newer arms BE8 (byte invariant) you would have to do even more work to try to make something generic that also tries to compensate for little intel, little arm, BE32 arm and BE8 arm. Xscale tends to run big endian natively but can be run as little endian to reduce the headaches. You may be assuming that because an ARM clone is big endian then all are big endian, another bad assumption.
I'm writing a Chip 8 emulator as an introduction to emulation and I'm kind of lost. Basically, I've read a Chip 8 ROM and stored it in a char array in memory. Then, following a guide, I use the following code to retrieve the opcode at the current program counter (pc):
// Fetch opcode
opcode = memory[pc] << 8 | memory[pc + 1];
Chip 8 opcodes are 2 bytes each. This is code from a guide which I vaguely understand as adding 8 extra bit spaces to memory[pc] (using << 8) and then merging memory[pc + 1] with it (using |) and storing the result in the opcode variable.
Now that I have the opcode isolated however, I don't really know what to do with it. I'm using this opcode table and I'm basically lost in regards to matching the hex opcodes I read to the opcode identifiers in that table. Also, I realize that many of the opcodes I'm reading also contain operands (I'm assuming the latter byte?), and that is probably further complicating my situation.
Help?!
Basically once you have the instruction you need to decode it. For example from your opcode table:
if ((inst&0xF000)==0x1000)
{
write_register(pc,(inst&0x0FFF)<<1);
}
And guessing that since you are accessing rom two bytes per instruction, the address is probably a (16 bit) word address not a byte address so I shifted it left one (you need to study how those instructions are encoded, the opcode table you provided is inadequate for that, well without having to make assumptions).
There is a lot more that has to happen and I dont know if I wrote anything about it in my github samples. I recommend you create a fetch function for fetching instructions at an address, a read memory function, a write memory function a read register function, write register function. I recommend your decode and execute function decodes and executes only one instruction at a time. Normal execution is to just call it in a loop, it provides the ability to do interrupts and things like that without a lot of extra work. It also modularizes your solution. By creating the fetch() read_mem_byte() read_mem_word() etc functions. You modularize your code (at a slight cost of performance), makes debugging much easier as you have a single place where you can watch registers or memory accesses and figure out what is or isnt going on.
Based on your question, and where you are in this process, I think the first thing you need to do before writing an emulator is to write a disassembler. Being a fixed instruction length instruction set (16 bits) that makes it much much easier. You can start at some interesting point in the rom, or at the beginning if you like, and decode everything you see. For example:
if ((inst&0xF000)==0x1000)
{
printf("jmp 0x%04X\n",(inst&0x0FFF)<<1);
}
With only 35 instructions that shouldnt take but an afternoon, maybe a whole saturday, being your first time decoding instructions (I assume that based on your question). The disassembler becomes the core decoder for your emulator. Replace the printf()s with emulation, even better leave the printfs and just add code to emulate the instruction execution, this way you can follow the execution. (same deal have a disassemble a single instruction function, call it for each instruction, this becomes the foundation for your emulator).
Your understanding needs to be more than vague as to what that fetch line of code is doing, in order to pull off this task you are going to have to have a strong understanding of bit manipulation.
Also I would call that line of code you provided buggy or at least risky. If memory[] is an array of bytes, the compiler might very well perform the left shift using byte sized math, resulting in a zero, then zero orred with the second byte results in only the second byte.
Basically a compiler is within its rights to turn this:
opcode = memory[pc] << 8) | memory[pc + 1];
Into this:
opcode = memory[pc + 1];
Which wont work for you at all, a very quick fix:
opcode = memory[pc + 0];
opcode <<= 8;
opcode |= memory[pc + 1];
Will save you some headaches. Minimal optimization will save the compiler from storing the intermediate results to ram for each operation resulting in the same (desired) output/performance.
The instruction set simulators I wrote and mentioned above are not intended for performance but instead readability, visibility, and hopefully educational. I would start with something like that then if performance for example is of interest you will have to re-write it. This chip8 emulator, once experienced, would be an afternoon task from scratch, so once you get through this the first time you could re-write it maybe three or four times in a weekend, not a monumental task (to have to re-write). (the thumbulator one took me a weekend, for the bulk of it. The msp430 one was probably more like an evening or two worth of work. Getting the overflow flag right, once and for all, was the biggest task, and that came later). Anyway, point being, look at things like the mame sources, most if not all of those instruction set simulators are designed for execution speed, many are barely readable without a fair amount of study. Often heavily table driven, sometimes lots of C programming tricks, etc. Start with something manageable, get it functioning properly, then worry about improving it for speed or size or portability or whatever. This chip8 thing looks to be graphics based so you are going to also have to deal with a lot of line drawing and other bit manipulation on a bitmap/screen/wherever. Or you could just call api or operating system functions. Basically this chip8 thing is not your traditional instruction set with registers and a laundry list of addressing modes and alu operations.
Basically -- Mask out the variable part of the opcode, and look for a match. Then use the variable part.
For example 1NNN is the jump. So:
int a = opcode & 0xF000;
int b = opcode & 0x0FFF;
if(a == 0x1000)
doJump(b);
Then the game is to make that code fast or small, or elegant, if you like. Good clean fun!
Different CPUs store values in memory differently. Big endian machines store a number like $FFCC in memory in that order FF,CC. Little-endian machines store the bytes in reverse order CC, FF (that is, with the "little end" first).
The CHIP-8 architecture is big endian, so the code you will run has the instructions and data written in big endian.
In your statement "opcode = memory[pc] << 8 | memory[pc + 1];", it doesn't matter if the host CPU (the CPU of your computer) is little endian or big endian. It will always put a 16-bit big endian value into an integer in the correct order.
There are a couple of resources that might help: http://www.emulator101.com gives a CHIP-8 emulator tutorial along with some general emulator techniques. This one is good too: http://www.multigesture.net/articles/how-to-write-an-emulator-chip-8-interpreter/
You're going to have to setup a bunch of different bit masks to get the actual opcode from the 16-bit word in combination with a finite state machine in order to interpret those opcodes since it appears that there are some complications in how the opcodes are encoded (i.e., certain opcodes have register identifiers, etc., while others are fairly straight-forward with a single identifier).
Your finite state machine can basically do the following:
Get the first nibble of the opcode using a mask like `0xF000. This will allow you to "categorize" the opcode
Based on the function category from step 1, apply more masks to either get the register values from the opcode, or whatever other variables might be encoded with the opcode that will narrow down the actual function that would need to be called, as well as it's arguments.
Once you have the opcode and the variable information, do a look-up into a fixed-length table of functions that have the appropriate handlers to coincide with the opcode functionality and the variables that go along with the opcode. While you can, in your state machine, hard-code the names of the functions that would go with each opcode once you've isolated the proper functionality, a table that you initialize with function-pointers for each opcode is a more flexible approach that will let you modify the code functionality easier (i.e., you could easily swap between debug handlers and "normal" handlers, etc.).