Sequential consistency of file IO (and other OS-related operations) on Linux - c++

I want to know if file IO operations by multiple processes/threads are guaranteed to be sequential consistent on Linux. And if not (as this thread says), how should I code to make sure they are sequential consistent? Consider the following example, where FILE_1 and FILE_2 are two distinct file names with absolute paths where both Process A and B have read-write access.
Process A first creates FILE_1 and then creates FILE_2:
FILE* fp1A = fopen(FILE_1, "w");
fclose(fp1A);
FILE* fp2A = fopen(FILE_2, "w");
fclose(fp2A);
Process B first reads FILE_2 and if success, reads FILE_1:
FILE* fp2B = fopen(FILE_2, "r");
if (fp2B != NULL) {
FILE* fp1B = fopen(FILE_1, "r");
// QUESTION: is fp1B guaranteed to be not NULL here?
}
Question is given by the comment above. In other words, if one process does some file IO operations in a given order specified by its source code, are all other processes going to see the effects of these operations on the system in the same order? Is this guaranteed by some standards (POSIX etc.) or implementation defined?
What if I change "file IO" to other operations which have some visible effect on the system in a broader sense (e.g. changing a kernel parameter)?
BACKGROUND: I have been studying memory ordering in the C++11 thread model. But those concepts only concerns memory rather than OS functionalities such as file IO. I understand this is because it is a language standard independent of OS. So I want to know if any other standards provide similar concepts for OS.

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Does my C++ code handle 100GB+ file copying? [closed]

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I need a cross-platform portable function that is able to copy a 100GB+ binary file to a new destination. My first solution was this:
void copy(const string &src, const string &dst)
{
FILE *f;
char *buf;
long len;
f = fopen(src.c_str(), "rb");
fseek(f, 0, SEEK_END);
len = ftell(f);
rewind(f);
buf = (char *) malloc((len+1) * sizeof(char));
fread(buf, len, 1, f);
fclose(f);
f = fopen(dst.c_str(), "a");
fwrite(buf, len, 1, f);
fclose(f);
}
Unfortunately, the program was very slow. I suspect the buffer had to keep 100GB+ in the memory. I'm tempted to try the new code (taken from Copy a file in a sane, safe and efficient way):
std::ifstream src_(src, std::ios::binary);
std::ofstream dst_ = std::ofstream(dst, std::ios::binary);
dst_ << src_.rdbuf();
src_.close();
dst_.close();
My question is about this line:
dst_ << src_.rdbuf();
What does the C++ standard say about it? Does the code compiled to byte-by-byte transfer or just whole-buffer transfer (like my first example)?
I'm curious does the << compiled to something useful for me? Maybe I don't have to invest my time on something else, and just let the compiler do the job inside the operator? If the operator translates to looping for me, why should I do it myself?
PS: std::filesystem::copy is impossible as the code has to work for C++11.
The crux of your question is what happens when you do this:
dst_ << src_.rdbuf();
Clearly this is two function calls: one to istream::rdbuf(), which simply returns a pointer to a streambuf, followed by one to ostream::operator<<(streambuf*), which is documented as follows:
After constructing and checking the sentry object, checks if sb is a null pointer. If it is, executes setstate(badbit) and exits. Otherwise, extracts characters from the input sequence controlled by sb and inserts them into *this until one of the following conditions are met: [...]
Reading this, the answer to your question is that copying a file in this way will not require buffering the entire file contents in memory--rather it will read a character at a time (perhaps with some chunked buffering, but that's an optimization that shouldn't change our analysis).
Here is one implementation: https://gcc.gnu.org/onlinedocs/libstdc++/libstdc++-api-4.6/a01075_source.html (__copy_streambufs). Essentially it a loop calling sgetc() and sputc() repeatedly until EOF is reached. The memory required is small and constant.
The C++ standard (I checked C++98, so this should be extremely compatible) says in [lib.ostream.inserters]:
basic_ostream<charT,traits>& operator<<
(basic_streambuf<charT,traits> *sb);
Effects: If sb is null calls setstate(badbit) (which may throw ios_base::failure).
Gets characters from sb and inserts them in *this. Characters are read from sb and inserted until any of the following occurs:
end-of-file occurs on the input sequence;
inserting in the output sequence fails (in which case the character to be inserted is not extracted);
an exception occurs while getting a character from sb.
If the function inserts no characters, it calls setstate(failbit) (which may throw ios_base::failure (27.4.4.3)). If an exception was thrown while extracting a character, the function set failbit in error state, and if failbit is on in exceptions() the caught exception is rethrown.
Returns: *this.
This description says << on rdbuf works on a character-by-character basis. In particular, if inserting of a character fails, that exact character remains unread in the input sequence. This implies that an implementation cannot just extract the whole contents into a single huge buffer upfront.
So yes, there's a loop somewhere in the internals of the standard library that does a byte-by-byte (well, charT really) transfer.
However, this does not mean that the whole thing is completely unbuffered. This is simply about what operator<< does internally. Your ostream object will still accumulate data internally until its buffer is full, then call write (or whatever low-level function your OS uses).
Unfortunately, the program was very slow.
Your first solution is wrong for a very simple reason: it reads the entire source file in memory, then write it entirely.
Files have been invented (perhaps in the 1960s) to handle data that don't fit in memory (and has to be in some "slower" storage, at that time hard disks or drums, or perhaps even tapes). And they have always been copied by "chunks".
The current (Unix-like) definition of file (as a sequence of bytes than is open-ed, read, write-n, close-d) is more recent than 1960s. Probably the late 1970s or early 1980s. And it comes with the notion of streams (which has been standardized in C with <stdio.h> and in C++ with std::fstream).
So your program has to work (like every file copying program today) for files much bigger than the available memory.You need some loop to read some buffer, write it, and repeat.
The size of the buffer is very important. If it is too small, you'll make too many IO operations (e.g. system calls). If it is too big, IO might be inefficient or even not work.
In practice, the buffer should today be much less than your RAM, typically several megabytes.
Your code is more C like than C++ like because it uses fopen. Here is a possible solution in C with <stdio.h>. If you code in genuine C++, adapt it to <fstream>:
void copyfile(const char*destpath, const char*srcpath) {
// experiment with various buffer size
#define MYBUFFERSIZE (4*1024*1024) /* four megabytes */
char* buf = malloc(MYBUFFERSIZE);
if (!buf) { perror("malloc buf"); exit(EXIT_FAILURE); };
FILE* filsrc = fopen(srcpath, "r");
if (!filsrc) { perror(srcpath); exit(EXIT_FAILURE); };
FILE* fildest = fopen(destpath, "w");
if (!fildest) { perror(destpath); exit(EXIT_FAILURE); };
for (;;) {
size_t rdsiz = fread(buf, 1, MYBUFFERSIZE, filsrc);
if (rdsiz==0) // end of file
break;
else if (rdsiz<0) // input error
{ perror("fread"); exit(EXIT_FAILURE); };
size_t wrsiz = fwrite(buf, rdsiz, 1, fildest);
if (wrsiz != 1) { perror("fwrite"); exit(EXIT_FAILURE); };
}
if (fclose(filsrc)) { perror("fclose source"); exit(EXIT_FAILURE); };
if (fclose(fildest)) { perror("fclose dest"); exit(EXIT_FAILURE); };
}
For simplicity, I am reading the buffer in byte components and writing it as a whole. A better solution is to handle partial writes.
Apparently dst_ << src_.rdbuf(); might do some loop internally (I have to admit I never used it and did not understand that at first; thanks to Melpopene for correcting me). But the actual buffer size matters a big lot. The two other answers (by John Swinck and by melpomene) focus on that rdbuf() thing. My answer focus on explaining why copying can be slow when you do it like in your first solution, and why you need to loop and why the buffer size matters a big lot.
If you really care about performance, you need to understand implementation details and operating system specific things. So read Operating systems: three easy pieces. Then understand how, on your particular operating system, the various buffering is done (there are several layers of buffers involved: your program buffers, the standard stream buffers, the kernel buffers, the page cache). Don't expect your C++ standard library to buffer in an optimal fashion.
Don't even dream of coding in standard C++ (without operating system specific stuff) an optimal or very fast copying function. If performance matters, you need to dive in OS specific details.
On Linux, you might use time(1), oprofile(1), perf(1) to measure your program's performance. You could use strace(1) to understand the various system calls involved (see syscalls(2) for a list). You might even code (in a Linux specific way) using directly the open(2), read(2), write(2), close(2) and perhaps readahead(2), mmap(2), posix_fadvise(2), madvise(2), sendfile(2) system calls.
At last, large file copying are limited by disk IO (which is the bottleneck). So even by spending days in optimizing OS specific code, you won't win much. The hardware is the limitation. You probably should code what is the most readable code for you (it might be that dst_ << src_.rdbuf(); thing which is looping) or use some library providing file copy. You might win a tiny amount of performance by tuning the various buffer sizes.
If the operator translates to looping for me, why should I do it myself?
Because you have no explicit guarantee on the actual buffering done (at various levels). As I explained, buffering matters for performance. Perhaps the actual performance is not that critical for you, and the ordinary settings of your system and standard library (and their default buffers sizes) might be enough.
PS. Your question contains at least 3 different questions (but related ones). I don't find it clear (so downvoted it), because I did not understand what is the most relevant one. Is it : performance? robustness? meaning of dst_ << src_.rdbuf();? Why is the first solution slow? How to copy large files quickly?

Do we need mutex to perform multithreading file IO

I'm trying to do random write (Benchmark test) to a file using multiple threads (pthread). Looks like if I comment out mutex lock the created file size is less than actual as if Some writes are getting lost (always in some multiple of chunk size). But if I keep the mutex it's always exact size.
Is my code have a problem in other place and mutex is not really required (as suggested by #evan ) or mutex is necessary here
void *DiskWorker(void *threadarg) {
FILE *theFile = fopen(fileToWrite, "a+");
....
for (long i = 0; i < noOfWrites; ++i) {
//pthread_mutex_lock (&mutexsum);
// For Random access
fseek ( theFile , randomArray[i] * chunkSize , SEEK_SET );
fputs ( data , theFile );
//Or for sequential access (in this case above 2 lines would not be here)
fprintf(theFile, "%s", data);
//sequential access end
fflush (theFile);
//pthread_mutex_unlock(&mutexsum);
}
.....
}
You are opening a file using "append mode". According to C11:
Opening a file with append mode ('a' as the first character in the
mode argument) causes all subsequent writes to the file to be forced
to the then current end-of-file, regardless of intervening calls to
the fseek function.
C standard does not specified how exactly this should be implemented, but on POSIX system this is usually implemented using O_APPEND flag of open function, while flushing data is done using function write. Note that fseek call in your code should have no effect.
I think POSIX requires this, as it describes how redirecting output in append mode (>>) is done by the shell:
Appended output redirection shall cause the file whose name results
from the expansion of word to be opened for output on the designated
file descriptor. The file is opened as if the open() function as
defined in the System Interfaces volume of POSIX.1-2008 was called
with the O_APPEND flag. If the file does not exist, it shall be
created.
And since most programs use FILE interface to send data to stdout, this probably requires fopen to use open with O_APPEND and write (and not functions like pwrite) when writing data.
So if on your system fopen with 'a' mode uses O_APPEND and flushing is done using write and your kernel and filesystem correctly implement O_APPEND flag, using mutex should have no effect as writes do not intervene:
If the O_APPEND flag of the file status flags is set, the file
offset shall be set to the end of the file prior to each write and no
intervening file modification operation shall occur between changing
the file offset and the write operation.
Note that not all filesystems support this behavior. Check this answer.
As for my answer to your previous question, my suggestion was to remove mutex as it should have no effect on the size of a file (and it didn't have any effect on my machine).
Personally, I never really used O_APPEND and would be hesitant to do so, as its behavior might not be supported at some level, plus its behavior is weird on Linux (see "bugs" section of pwrite).
You definitely need a mutex because you are issuing several different file commands. The underlying file subsystem can't possibly know how many file commands you are going to call to complete your whole operation.
So you need the mutex.
In your situation you may find you get better performance putting the mutex outside the loop. The reason being that, otherwise, switching between threads may cause excessive skipping between different parts of the disk. Hard disks take about 10ms to move the read/write head so that could potentially slow things down a lot.
So it might be a good idea to benchmark that.

Atomic writing to file on linux

Is there a way to dump a buffer to file atomically?
By "atomically" I mean: if for example someone terminates my application during writing, I'd like to have file in either before- or after-writing state, but not in a corrupted intermediate state.
If the answer is "no", then probably it could be done with a really small buffers?
For example, can I dump 2 consequent int32_t variables with a single 8 bytes fwrite (on x64 platform), and be sure that both of those int32s are dumped, or neither of them, but not only just one of them?
I recommend writing to a temporary file and then doing a rename(2) on it.
ofstream o("file.tmp"); //Write to a temporary file
o << "my data";
o.close();
//Perform an atomic move operation... needed so readers can't open a partially written file
rename("file.tmp", "file.real");

Parallel IO & Append

When I run my small-scale parallel codes, I typically output N files (N being number of processors) in the form fileout.dat.xxx where xxx is the processor number (using I3.3) and then just cat them into a single fileout.dat file after the code is finished.
My question is can I use ACCESS='append' or POSITION='append' in the OPEN statement and have all processors write to the same file?
In practice, no. POSITION='append' merely says that the file pointer will be at the end of file after the open statement is executed. It is, however, possible to change the file position, e.g. with the BACKSPACE, REWIND or such statements. Thus, Fortran POSITION='append' does not correspond to the POSIX O_APPEND, and hence a POSIX OS cannot ensure that all writes only append to the file and do not overwrite older data.
Furhtermore, in case you run your code on a cluster, be aware that O_APPEND does not work on many networked file systems such as NFS.
In order to do parallel I/O with several processes/threads writing to a single file, use ACCESS='direct' or ACCESS='stream' and have the processes agree on which records/byte ranges to write to.

is pwrite after dup race safe?

On Linux pwrite operation (which is seek+write) is atomic, meaning doing pwrite-s in multiple threads with one file descriptor is safe.
I want to create file descriptor duplicate, using dup(). Now, having fd1 and fd2 - will pwrite-s work as expected, or there's danger of race condition?
File descriptor pairs created through dup share the same file status, (e.g. an lseek operation on one file descriptor will affect the other), because they refer to the same entry in the process open files table, which means they are essentially indistinguishable. The only thing they do not have in common is file descriptor flags, (e.g. FD_CLOEXEC.)
From the man page:
After a successful return from dup()
or dup2(), the old and new file
descriptors may be used
interchangeably. They refer to the
same open file description (see
open(2)) and thus share file offset
and file status flags; for example, if
the file offset is modified by using
lseek(2) on one of the descriptors,
the offset is also changed for the
other.
Given that dup allows you to use the two file descriptors interchangeably, (because they refer to the same file in the process file table) I assume this implies that calling pwrite on one would be the same as calling it on the other, and thus be atomic.
I think pwrite is an atomic operation if the number of bytes you're writing is less than PIPE_BUF of the pipe you're writing to (from the POSIX programmer's manual).