I am going through Mark Burgess's "The GNU C Programming Tutorial". I have come across the following information:
Even though low-level fle routines do not use buffering, and once you call write, your data can be read from the file immediately, it may take up to a minute before your data is physically written to disk. (Page:142)
Firstly, is "it may take up to a minute(some time) before your data is written to disk" true?
Secondly, when low level file routines are not using buffering why will the delay take place?
There are two places where I/O buffering can occur (at least — it could be more than just two).
One is in the application; the standard I/O functions using FILE * use buffered I/O unless you use setvbuf() to prevent it.
The other is in the kernel. Disk I/O normally goes into the kernel buffer pool, and eventually gets written by the kernel to disk. There are ways around that (O_DIRECT on Linux; raw devices on classic Unix; etc). The key point is that the write() system call normally writes to he kernel buffer pool. The kernel takes responsibility for ensuring that the data is written to disk safely and correctly (journalling, …).
The kernel doesn't write everything to disk immediately because (a) you may add more changes to the data, (b) other people may need to read or write the data, (c) the disk drive may be busy writing something else at the other end of its 1 TiB of storage and it will take time to get the write head in position to take your data, and it would be better for the overall performance of the system if it scheduled other work before writing your changed buffer to disk. It will get written to disk. It is just not defined when, and it could be fractions of a second or multiple seconds or longer, though most often it will not take minutes for the data to be written to disk.
These days, there could also be buffering in the RAID controllers, and maybe in the individual disks inside the RAID setup, and maybe there's network buffering too if it is a remotely-mounted file system. Those add extra levels of buffering.
The read() and write() and related low-level I/O functions do not have any client-side (application) buffering — unlike the standard C I/O functions.
A file is said to be buffered, when its contents are not outputted or inputted directly. Instead, the file's bytes are written to a temporary buffer in memory.
For example, if you are reading from a file, you are reading from the buffer. Once you have read all the characters in the buffer, it is replenished with new bytes from the file. The reason for this indirectness, is that a memory read is much faster than a hard disk read.
The calls read and write are low-level, and do not perform buffering. The stdio.h calls like getc and putc, do use buffering. These higher-level APIs only call the low level ones, when the buffer must be replenished.
Writing to the hard drive is much slower than writing to RAM. When you write to a drive it writes to memory, but doesn't always write to the disk immediately. The data might not be written to disk until that part of memory needs to be overwritten to make room for something else. This is called a Write-Back cache.
Related
I was reading about sendfile(2) recently, and the man page states:
sendfile() copies data between one file descriptor and another.
Because this copying is done within the kernel, sendfile() is more
efficient than the combination of read(2) and write(2), which would
require transferring data to and from user space.
It got me thinking, why exactly is the combination of read()/write() slower? The man page focuses on extra copying that has to happen to and from userspace, not the total number of calls required. I took a short look at the kernel code for read and write but didn't see the copy.
Why does the copy exist in the first place? Couldn't the kernel just read from the passed buffer on a write() without first copying the whole thing into kernel space?
What about asynchronous IO interfaces like AIO and io_uring? Do they also copy?
why exactly is the combination of read()/write() slower?
The manual page is quite clear about this. Doing read() and then write() requires to copy the data two times.
Why does the copy exist in the first place?
It should be quite obvious: since you invoke read, you want the data to be copied to the memory of your process, in the specified destination buffer. Same goes for write: you want the data to be copied from the memory of your process. The kernel doesn't really know that you just want to do a read + write, and that copying back and forth two times could be avoided.
When executing read, the data is copied by the kernel from the file descriptor to the process memory. When executing write the data is copied by the kernel from the process memory to the file descriptor.
Couldn't the kernel just read from the passed buffer on a write() without first copying the whole thing into kernel space?
The crucial point here is that when you read or write a file, the file has to be mapped from disk to memory by the kernel in order for it to be read or written. This is called memory-mapped file I/O, and it's a huge factor in the performance of modern operating systems.
The file content is already present in kernel memory, mapped as a memory page (or more). In case of a read, the data needs to be copied from that file kernel memory page to the process memory, while in case of a write, the data needs to be copied from the process memory to the file kernel memory page. The kernel will then ensure that the data in the kernel memory page(s) corresponding to the file is correctly written back to disk when needed (if needed at all).
This "intermediate" kernel mapping can be avoided, and the file mapped directly into userspace memory, but then the application would have to manage it manually, which is complicated and easy to mess up. This is why, for normal file operations, files are mapped into kernel memory. The kernel provides high level APIs for userspace programs to interact with them, and the hard work is left to the kernel itself.
The sendfile syscall is much faster because you do not need to perform the copy two times, but only once. Assuming that you want to do a sendfile of file A to file B, then all the kernel needs to do is to copy the data from A to B. However, in the case of read + write, the kernel needs to first copy from A to your process, and then from your process to B. This double copy is of course slower, and if you don't really need to read or manipulate the data, then it's a complete waste of time.
FYI, sendfile itself is basically an easy-to-use wrapper around splice (as can bee seen from the source code), which is a more generic syscall to perform zero-copy data transfer between file descriptors.
I took a short look at the kernel code for read and write but didn't see the copy.
In terms of kernel code, the whole process for reading a file is very complicated, but what the kernel ends up doing is a "special" version of memcpy(), called copy_to_user(), which copies the content of the file from the kernel memory to the userspace memory (doing the appropriate checks before performing the actual copy). More specifically, for files, the copyout() function is used, but the behavior is very similar, both end up calling raw_copy_to_user() (which is architecture-dependent).
What about asynchronous IO interfaces like AIO and io_uring? Do they also copy?
The aio_{read,write} libc functions defined by POSIX are just asynchronous wrappers around read and write (i.e. they still use read and write under the hood). These still copy data to/from userspace.
io_uring can provide zero-copy operations, when using the O_DIRECT flag of open (see the manual page):
O_DIRECT (since Linux 2.4.10)
Try to minimize cache effects of the I/O to and from this
file. In general this will degrade performance, but it is
useful in special situations, such as when applications do
their own caching. File I/O is done directly to/from user-
space buffers. The O_DIRECT flag on its own makes an effort
to transfer data synchronously, but does not give the
guarantees of the O_SYNC flag that data and necessary metadata
are transferred. To guarantee synchronous I/O, O_SYNC must be
used in addition to O_DIRECT. See NOTES below for further
discussion.
This should be done carefully though, as it could very well degrade performance in case the userspace application does not do the appropriate caching on its own (if needed).
See also this related detailed answer on asynchronous I/O, and this LWN article on io_uring.
I keep on reading that fread() and fwrite() are buffered library calls. In case of fwrite(), I understood that once we write to the file, it won't be written to the hard disk, it will fill the internal buffer and once the buffer is full, it will call write() system call to write the data actually to the file.
But I am not able to understand how this buffering works in case of fread(). Does buffered in case of fread() mean, once we call fread(), it will read more data than we originally asked and that extra data will be stored in buffer (so that when 2nd fread() occurs, it can directly give it from buffer instead of going to hard disk)?
And I have following queries also.
If fread() works as I mention above, then will first fread() call read the data that is equal to the size of the internal buffer? If that is the case, if my fread() call ask for more bytes than internal buffer size, what will happen?
If fread() works as I mention above, that means at least one read() system call to kernel will happen for sure in case of fread(). But in case of fwrite(), if we only call fwrite() once during the program execution, we can't say for sure that write() system call be called. Is my understanding correct?
Will the internal buffer be maintained by OS?
Does fclose() flush the internal buffer?
There is buffering or caching at many different levels in a modern system. This might be typical:
C standard library
OS kernel
disk controller (esp. if using hardware RAID)
disk drive
When you use fread(), it may request 8 KB or so if you asked for less. This will be stored in user-space so there is no system call and context switch on the next sequential read.
The kernel may read ahead also; there are library functions to give it hints on how to do this for your particular application. The OS cache could be gigabytes in size since it uses main memory.
The disk controller may read ahead too, and could have a cache size up to hundreds of megabytes on smallish systems. It can't do as much in terms of read-ahead, because it doesn't know where the next logical block is for the current file (indeed it doesn't even know what file it is reading).
Finally, the disk drive itself has a cache, perhaps 16 MB or so. Like the controller, it doesn't know what file it is reading. For many years one disk block was 512 bytes, but it got a little larger (a few KB) recently with multi-terabyte disks.
When you call fclose(), it will probably deallocate the user-space buffer, but not the others.
Your understanding is correct. And any buffered fwrite data will be flushed when the FILE* is closed. The buffered I/O is mostly transparent for I/O on regular files.
But for terminals and other character devices you may care. Another instance where buffered I/O may be an issue is if you read from the file that one process is writing to from another process -- a common example is if a program writes text to a log file during operation, and the user runs a command like tail -f program.log to watch the content of the log file live. If the writing process has buffering enabled and it doesn't explicitly flush the log file, it will make it difficult to monitor the log file.
What really happens when write() system call is executed?
Lets say I have a program which writes certain data into a file using write() function call. Now C library has its own internal buffer and OS too has its own buffer.
What interaction takes place between these buffers ?
Is it like when C library buffer gets filled completely, it writes to OS buffer and when OS buffer gets filled completely, then the actual write is done on the file?
I am looking for some detailed answers, useful links would also help. Consider this question for a UNIX system.
The write() system call (in fact all system calls) are nothing more that a contract between the application program and the OS.
for "normal" files, the write() only puts the data on a buffer, and marks that buffer as "dirty"
at some time in the future, these dirty buffers will be collected and actually written to disk. This can be forced by fsync()
this is done by the .write() "method" in the mounted-filesystem-table
and this will invoke the hardware's .write() method. (which could involve another level of buffering, such as DMA)
modern hard disks have there own buffers, which may or may not have actually been written to the physical disk, even if the OS->controller told them to.
Now, some (abnormal) files don't have a write() method to support them. Imagine open()ing "/dev/null", and write()ing a buffer to it. The system could choose not to buffer it, since it will never be written anyway.
Also note that the behaviour of write() does depend on the nature of the file; for network sockets the write(fd,buff,size) can return before size bytes have been sent(write will return the number of characters sent). But it is impossible to find out where they are once they have been sent. They could still be in a network buffer (eg waiting for Nagle ...), or a buffer inside the network interface, or a buffer in a router or switch somewhere on the wire.
As far as I know...
The write() function is a lower level thing where the library doesn't buffer data (unlike fwrite() where the library does/may buffer data).
Despite that, the only guarantee is that the OS transfers the data to disk drive before the next fsync() completes. However, hard disk drives usually have their own internal buffers that are (sometimes) beyond the OS's control, so even if a subsequent fsync() has completed it's possible for a power failure or something to occur before the data is actually written from the disk drive's internal buffer to the disk's physical media.
Essentially, if you really must make sure that your data is actually written to the disk's physical media; then you need to redesign your code to avoid this requirement, or accept a (small) risk of failure, or ensure the hardware is capable of it (e.g. get a UPS).
write() writes data to operating system, making it visible for all processes (if it is something which can be read by other processes). How operating system buffers it, or when it gets written permanently to disk, that is very library, OS, system configuration and file system specific. However, sync() can be used to force buffers to be flushed.
What is quaranteed, is that POSIX requires that, on a POSIX-compliant file system, a read() which can be proved to occur after a write() has returned must return the written data.
OS dependant, see man 2 sync and (on Linux) the discussion in man 8 sync.
Years ago operating systems were supposed to implement an 'elevator algorithm' to schedule writes to disk. The idea would be to minimize the disk writing head movement, which would allow a good throughput for several processes accessing the disk at the same time.
Since you're asking for UNIX, you must keep in mind that a file might actually be on an FTP server, which you have mounted, as an example. For example files /dev and /proc are not files on the HDD, as well.
Also, on Linux data is not written to the hard drive directly, instead there is a polling process, that flushes all pending writes every so often.
But again, those are implementation details, that really don't affect anything from the point of view of your program.
I know that when you call fwrite or fprintf or rather any other function that writes to a file, the contents aren't immediately flushed to the disk, but buffered in the memory.
Firstly, where do the OS manage these buffers and how. Secondly, if you do the write to a file and later read in the content you wrote and assuming that the OS didn't flushed the contents between the time you wrote and read, how it knows that it has to return the read from the buffer? How does it handle this situation.
The reason I want to know this is that I'm interested in implementing my own buffering scheme in user-space, rather than kernel space as done by OS. That is, write to a file would be buffered in user-space and the actual write will only occur at a certain point. Consquently I also need to handle situations where read is called for the content that is still in the buffer. Is it possible to do all this in user-space.
Firstly, where do the OS manage these buffers and how
The functions fwrite and fprintf use stdio buffers which already are completely in userspace. The buffers are (likely) static arrays or perhaps malloced memory.
how it knows that it has to return the read from the buffer
It doesn't, so the changes aren't seen. Nothing actually happens to a file until the underlying system call (write) is called (and even then - read on).
Is it possible to do all this in user-space
No, it's not possible. The good news is that the kernel already has buffers so every write you do isn't atually translated into an actual write to the file. It is postponed and executed later. If in the meantime somebody tries to read from the file, the kernel is smart enough to serve him from the buffer.
Bits from TLPI:
When working with disk files, the read() and write() system calls
don’t directly ini- tiate disk access. Instead, they simply copy data
between a user-space buffer and a buffer in the kernel buffer cache.
When performing I/O on a disk file, a successful return from write()
doesn’t guarantee that the data has been transferred to disk, because
the kernel performs buffering of disk I/O in order to reduce disk
activity and expedite write() calls.
At some later point, the kernel writes (flushes) its buffer to the
disk.
If, in the interim, another process attempts to read these bytes of
the file, then the kernel automatically supplies the data from the
buffer cache, rather than from (the outdated contents of) the file.
So you may want to find out about sync and fsync.
Multiple levels of buffering are generally bad. The reason stdio buffers are useful is that they minimize the number of system calls performed. If a system call would be cheaper nobody would use stdio buffers anymore.
After providing the same program which reads a random generated input file and echoes the same string it read to an output. The only difference is that on one side I'm providing the read and write methods from linux syscalls, and on the other side I'm using fread/fwrite.
Timing my application with an input of 10Mb in size and echoing it to /dev/null, and making sure the file is not cached, I've found that libc's fwrite is faster by a LARGE scale when using very small buffers (1 byte in case).
Here is my output from time, using fwrite:
real 0m0.948s
user 0m0.780s
sys 0m0.012s
And using the syscall write:
real 0m8.607s
user 0m0.972s
sys 0m7.624s
The only possibility that I can think of is that internally libc is already buffering my input... Unfortunately I couldn't find that much information around the web, so maybe the gurus here could help me out.
Timing my application with an input of
10Mb in size and echoing it to
/dev/null, and making sure the file in
not cached, I've found that libc's
frwite is faster by a LARGE scale when
using very small buffers (1 byte in
case).
fwrite works on streams, which are buffered. Therefore many small buffers will be faster because it won't run a costly system call until the buffer fills up (or you flush it or close the stream). On the other hand, small buffers being sent to write will run a costly system call for each buffer - that's where you're losing the speed. With a 1024 byte stream buffer, and writing 1 byte buffers, you're looking at 1024 write calls for each kilobyte, rather than 1024 fwrite calls turning into one write - see the difference?
For big buffers the difference will be small, because there will be less buffering, and therefore a more consistent number of system calls between fwrite and write.
In other words, fwrite(3) is just a library routine that collects up output into chunks, and then calls write(2). Now, write(2), is a system call which traps into the kernel. That's where the I/O actually happens. There is some overhead for simply calling into the kernel, and then there is the time it takes to actually write something. If you use large buffers, you will find that write(2) is faster because it eventually has to be called anyway, and if you are writing one or more times per fwrite then the fwrite buffering overhead is just that: more overhead.
If you want to read more about it, you can have a look at this document, which explains standard I/O streams.
write(2) is the fundamental kernel operation.
fwrite(3) is a library function that adds buffering on top of write(2).
For small (e.g., line-at-a-time) byte counts, fwrite(3) is faster, because of the overhead for just doing a kernel call.
For large (block I/O) byte counts, write(2) is faster, because it doesn't bother with buffering and you have to call the kernel in both cases.
If you look at the source to cp(1), you won't see any buffering.
Finally, there is one last consideration: ISO C vs Posix. The buffered library functions like fwrite are specified in ISO C whereas kernel calls like write are Posix. While many systems claim Posix-compatibility, especially when trying to qualify for government contracts, in practice it's specific to Unix-like systems. So, the buffered ops are more portable. As a result, a Linux cp will certainly use write but a C program that has to work cross-platform may have to use fwrite.
You can also disable buffering with setbuf() function. When the buffering is disabled, fwrite() will be as slow as write() if not slower.
More information on this subject can be found there: http://www.gnu.org/s/libc/manual/html_node/Controlling-Buffering.html