I am recently studying file system in linux. I learned that when we call fopen(), the library call will call malloc() to allocate space for the FILE structure, and inside this FILE structure there will be a buffer for I/0. But later on I found that the write system call actually writes the data to the kernel buffer, So what's the difference between these two buffers?
You have to understand two things: fwrite() is a standard library's routine operating on FILE structure, but write() is a system call. I bet fwrite() uses write() internally. Nothing keeps fwrite() from providing user space IO-buffering until it is ready to pass your data on to the write() syscall.
write() syscall in it's turn goes straight to the kernel and says: "Hey kernel, I've got this user space buffer here. Will you write this one to the storage for me?". And here it's up to the kernel what to do next: it will either go directly to storage to write the data, or, most likely, copy the data to kernel buffer, until it decides it's time to modify storage data.
Turning back to your question. Any kind of buffering is done to accumulate data in order to postpone turning to more expensive operations: standard library may consider invoking syscall on every len-byte expensive, kernel considers going to hard disk on every syscall expensive and so on.
You might want to read this to see how far buffering goes https://fgiesen.wordpress.com/2015/10/25/reading-and-writing-are-less-symmetric-than-you-probably-think/
The FILE structure holds the meta data about the opened file (mode, stream position, etc). It is part of the C Standard I/O interface.
The buffer allocated as part of FILE takes only a limited amount of data (e.g. when the stream is buffered). It is deallocated upon fclose(). You may even provide your own user space stdio buffer with setvbuf().
The kernel buffer receives the file contents written by write(), whenever the stream is flushed or the associated file descriptor is closed.
The FILE structure holds the information about the opened file.this defines the FILE struct members. But in kernel level a file has been accessed by inode, buffer cache.
Data is getting read/write to disk from user space through buffer cache using method copy_to_user and copy_from_user.
There is a very big difference between the two buffers, one is kernel buffer and the other is the user buffer. So, what basically happens when you do a I/O is that the buffer from the user space is copied into a buffer in the kernel space. The function copy_from_user() does this task.
Now the question which arises is why do we need two buffers, when the kernel has access into the user space? The reason is that the kernel does not want to read the user buffer directly because both the kernel and the user space has different address spaces and so a valid address in user space might not be a valid address in kernel.
In kernel if a non-valid address is accessed then the system will panic immediately, so the function copy_from_user does the task of mapping the user space address and the kernel space address and checks if the address is accessible or not. If not then it simply returns EFAULT (bad address).
Related
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.
Assuming I understand the flow correctly, one would like to read few byes off an opened FILE stream, lets says, using fread:
the read syscall will copy the data from the kernel to the user space buffer
user space buffer (either allocated by glibc or provided by setvbuf...) will be copied to the buffer provided to fread
why is the 2nd step needed? why can I get a pointer to the user space buffer which I will decide if I want to store (copy) or not?
Thanks,
The purpose of the 2nd buffer is to amortize the system call overhead. If you read/write only a few bytes at a time, this second user space buffer will improve performance enormously. OTOH if you read/write a large chunk, the 2nd buffer can be bypassed, so you don't pay the price for double copying.
The second step is what it is all about. Kernel must take care of such operations. The api you use will be feeded afterwards with the result. This is usual kernelspace/user space behaviour. Read about it. You perhaps might not know it NOW but kernel space/ user space differentiation are basics of os infrastructure.
I am reading about Memory Mapped Files, the souce says it is faster than traditional methods to open a file or read a file such as an open system call and read system call respectively without giving the description that how open or read system call works.
So here's my question how the open system call works?
As far i know it will load the file into the memory, whereas by using mapped file only their addresses will be saved in the memory and when needed the requested page may be brought into the memory.
I expect clarification over my so far understanding.
EDIT
My previous understanding written above is almost wrong, for coorrect explanation refer to the accepted answer by Pawel.
Since you gave no details I'm assuming you are interested in behavior of Unix-like systems.
Actually open() system call only creates a file descriptor which then may be used by either mmap() or read().
Both memory mapped I/O and standard I/O internally access files on disk through page cache, a buffer in which files are cached in order to reduce number of I/O operations.
Standard I/O approach (using write() and read()) involves performing a system call which then copies data from (or to if you are writing) page cache to a buffer chosen by application. In addition to that non-sequential access requires another system call lseek(). System calls are expensive and so is copying data.
When a file is memory mapped usually a memory region in process address space is mapped directly to page cache, so that all reads and writes of already loaded data can be performed without any additional delay (no system calls, no data copying). Only when an application attempts to access file region that is not already loaded a page fault occurs and the kernel loads required data (whole page) from disk.
EDIT:
I see that I also have to explain memory paging. On most modern architectures there is physical memory which is a real piece of hardware and virtual memory which creates address spaces for processes. Kernel decides how addresses in virtual memory are mapped to addresses in physical memory. The smallest unit is a memory page (usually, but not always 4K). It does not have to be 1:1 mapping, for example all virtual memory pages may be mapped to the same physical address.
In memory mapped I/O part of application address space and kernel's page cache are mapped to the same physical memory region, hence program is able to directly access page cache.
Pawel has beautifully explained how read/writes are performed. Let me explain the original question: How does fopen(3) works:
when user space process encounters fopen(defined in libc or any user space library), it translates it into open(2) system call. First, it takes arguments from fopen, writes them into architecture specific registers along with open() syscall number. This number tells kernel the system call user space program wants to run. After loading these register, user space process interrupts kernel(via softirq, traditionally INT 80H on x86) and blocks.
Kernel verifies the arguments provided and access permissions etc, and then either returns error or invokes actual system call which is vfs_open() in this case. vfs_open() checks for available file descriptor in fd array and allocates struct file. The ref counts of accessed file is increased and fd is returned to user program. That's completes the working of open, and of most of the system calls in general.
open() together with read()/write(), followed by close() is undoubtedly much lengthy process than having memory mapped file in buffer cache.
For a lucid explanation of how open and read work on Linux, you can read this. The code snippets are from an older version of the kernel but the theory still holds.
You would still need to use the open() system call to get a valid file descriptor, which you would pass on to mmap(). As to why mmaped IO is faster, it is because there is no copy of data from (to) user space to (from) kernel space buffers which is what happens with read and write system calls.
I am using fopen/fread/fwrite/fseek on linux with gcc. is it necessary to allocate a memory buffer and use fread to read data sequentially into buffer before using the data?
When you use fread or the other file I/O functions in the C standard library, memory is buffered in several places.
Your application allocates a buffer which gets passed to fread. fread copies data into your buffer, and then you can do what you want with it. You are responsible for allocation/deallocation of this buffer.
The C library will usually create a buffer for every FILE* you have open. Data is read into this buffers in large chunks. This allows fread to satisfy many small requests without having to make a large number of system calls, which are expensive. This is what people mean when they say fread is buffered.
The kernel will also buffer files that are being read in the disk cache. This reduces the time needed for the read system call, since if data is already in memory, your program won't have to wait while the kernel fetches it from the disk. The kernel will hold on to recently read files, and it may read ahead for files which are being accessed sequentially.
The C library buffer is allocated automatically when you open a file and freed when you close the file. You don't have to manage it at all.
The kernel disk cache is stored in physical memory that isn't being used for anything else. Again, you don't have to manage this. The memory will be freed as soon as it's needed for something else.
You must pass a buffer (a buffer created by your code, malloced or local) to fread to pass the read data back to you. I don't know what do you exactly mean by saying "fread is buffered". Most 'C' library calls operate in this fashion. They will not return their internal storage (buffer or otherwise) to you and if they do, they will provide you a corresponding free/release functions.
Refer http://pubs.opengroup.org/onlinepubs/000095399/functions/fread.html It has a very basic example also.
With fread, yes, you have to allocate memory in your process and the system call will copy the data into your buffer.
In some specialised cases, you can handle data without copying it into userspace. See the sendfile system call, which copies data from one file descriptor to another directly. This can be used to transfer data from a file to a network socket without excessive copying.
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.