I'm trying to write to stdin and read from stdout ( and stderr ) from an external program, without changing the code.
I've tried using named pipes, but stdout doesn't show until the program is terminated and stdin only works on the first input( then cin is null ).
i've tried using /proc/[pid]/fd but that only writes and reads from the terminal and not the program.
i've tried writing a character device file for this and it worked, but only one program at a time ( this needs to work for multiple programs at a time ).
at this point, to my knowledge, I could write the driver that worked to multiplex the io across multiple programs but I don't think that's the "right" solution.
the main purpose of this is to view a feed of a program through a web interface. I'm sure there has to be someway to do this. is there anything I haven't tried that's been done before?
The typical way of doing this is:
Create anonymous pipes (not named pipes) with the pipe(2) system call for the new process's standard streams
Call fork(2) to spawn the child process
close(2) the appropriate ends of the pipes in both the parent and the child (e.g. for the stdin pipe, close the read end in the parent and close the write end in the child; vice-versa for the stdout and stderr pipes)
Use dup2(2) in the child to copy the pipe file descriptors onto file descriptors 0, 1, and 2, and then close(2) the remaining old descriptors
exec(3) the external application in the child process
In the parent process, simultaneously write to the child's stdin pipe and read from the child's stdout and stderr pipes. However, depending on how the child behaves, this can easily lead to deadlock if you're not careful. One way to avoid deadlock is to spawn separate threads to handle each of the 3 streams; another way is to use the select(2) system call to wait until one of the streams can be read from/written to without blocking, and then process that stream.
Even if you do this all correctly, you may still not see your program's output right away. This is typically due to buffering stdout. Normally, when stdout is going to a terminal, it's line-buffered—it gets flushed after every newline gets written. But when stdout is a pipe (or anything else that's not a terminal, like a file or a socket), it's fully buffered, and it only gets written to when the program has outputted a full buffer's worth of data (e.g. 4 KB).
Many programs have command line options to change their buffering behavior. For example, grep(1) has the --line-buffered flag to force it to line-buffer its output even when stdout isn't a terminal. If your external program has such an option, you should probably use it. If not, it's still possible to change the buffering behavior, but you have to use some sneaky tricks—see this question and this question for how to do that.
Related
My problem is a bit hard to explain properly as I do not understand fully the behavior behind it.
I have been working on pipe and pipelines in C, and I noticed some behavior that is a bit mysterious to me.
Let's take a few example: Let's try to pipe yes with head. (yes | head). Even though I coded the behavior in a custom program, I don't understand how the pipe knows when to stop piping ? It seems two underlying phenomenons are causing this (maybe), the SIGPIPE and/or the internal size a pipe can take. How does the pipe stop piping, is it when it's full ? But the size of a pipe is way superior to 10 "yes\n" no ? And SIGPIPE only works when the end read/write is closed no ?
Also let's take another example, for example cat and ls: cat | ls or even cat | cat | ls.
It seems the stdin of the pipe is waiting for input, but how does it know when to stop, i.e. after one input ? What are the mechanism that permits this behavior?
Also can anyone provide me with others examples of these very specific behavior if there are any in pipes and pipelines so I can get an good overview of theses mechanism ?
In my own implementation, I managed to replicate that behavior using waitpid. However how does the child process itself know when to stop ? Is it command specific ?
The write operation will block when the pipe buffer is full, the read operation will block when the buffer is empty.
When the write end of the pipe is closed, the reading process will get an EOF indication after reading all data from the buffer. Many programs will terminate in this case.
When the read end of the pipe is closed, the writing process will get a SIGPIPE. This will also terminate most programs.
When you run cat | ls, STDOUT of cat is connected to STDIN of ls, but ls does not read from STDIN. On the system where I checked this, ls simply ignores STDIN and the file descriptor will be closed when ls terminates.
You will see the output of ls, and cat will be waiting for input.
cat will not write anything to STDOUT before it has read enough data from STDIN, so it will not notice that the other end of the pipe has been closed.
cat will terminate when it detects EOF on STDIN which can be done by pressing CTRL+D or by redirecting STDIN from /dev/null, or when it gets SIGPIPE after trying to write to the pipe which will happen when you (type something and) press ENTER.
You can see the behavior with strace.
cat terminates after EOF on input which is shown as read(0, ...) returning 0.
strace cat < /dev/null | ls
cat killed by SIGPIPE.
strace cat < /dev/zero | ls
How does the pipe stop piping
The pipe stops piping when either end is closed.
If the input(write) end of the pipe is closed, then any data in the pipe is held until it is read from the output end. Once the buffer is emptied, anyone subsequently reading from the output end will get an EOF.
If the output(read) end of the pipe is closed, any data in the pipe will be discarded. Anyone subsequently writing to the input end will get a SIGPIPE/EPIPE. Note that a process merely holding open the input but not actively writing to it will not be signalled.
So when you type cat | ls you get a cat program with stdout connected to the input of the pipe and ls with stdin connected to the output. ls runs and outputs some stuff (to its stdout, which is still the terminal) and never reads from stdin. Once done it exits and closes the output of the pipe. Meanwhile cat is waiting for input from its stdin (the terminal). When it gets it (you type a line), it writes it to stdout, gets a SIGPIPE/EPIPE and exits (discarding the data as there's noone to write it to.) This closes the input of the pipe, so the pipe goes away now that both ends have been closed.
Now lets look at what happens with cat | cat | ls. You now have two pipes and two cat programs. As before ls runs and exits, closing the output of the second pipe. Now you type a line and the first cat reads it and copies it to the first pipe (still fully open) where the second cat reads it and copies it to the second pipe (which has its output closed), so it (the second cat) gets a SIGPIPE/EPIPE and exits (which closes the output of the first pipe). At this point the first cat is still waiting for input, so if you type a second line, it copies that to the now closed first pipe and gets a SIGPIPE/EPIPE and exits
How does the pipe stop piping, is it when it's full ?
A pipe has several states:
if you obtain the pipe through a call to pipe(2) (an unnamed pipe) both file descriptors are already open, so this doesn't apply to it (you start in point 2. below). When you open a named pipe, your open(2) call (depending if you have open with O_READ, O_WRITE, or O_RDWR. The pipe has two sides, the writer and the reader side. When you open it, you attach to the sides, depending on how do you open it. Well, up to here, the pipe blocks any open(2) call, until both sides have at least one process tied to them. So, if you open a pipe and read(2) from it, then your open will be blocked, until other process has opened it to read.
once both extremes have it open, the readers (the process issuing a read(2) call) block when the pipe is empty, and the writers (the processes issuing a write(2) call) block whenever the write call cannot be satisfied due to fillin completely the pipe. Old implementations of pipes used the filesystem to hold the data, and the data was stored only in the direct addressed disk blocks. This meant (as there are 10 such blocks in an inode) that you normally had space in the pipe to hold 10 blocks, after that, the writers are blocked. Later, pipes were implemented using the socket infrastructure in BSD systems, which allowed you to control the buffer size with ioctl(2) calls. Today, IMHO, pipes use a common implementation, that is separate from sockets also.
When the processes close the pipe continues to work as said in point 2. above, until the number of readers/writers collapses to zero. At that point, the pipe starts giving End Of File condition to all readers (this means read(2) syscall will return 0 bytes, without blocking) and error (cannot write to pipe) to writers. In addition, the kernel sends a signal (which normally aborts the writer processes) SIGPIPE to every process that has the pipe open for writing. If you have not ignored that signal or you have not installed a signal handler for it, your process will die. In this state, it's impossible to reopen the pipe again, until all processes have closed it.
A common error is when you pipe() or you open a pipe with O_RDWR, and the other process closes its file descriptor, and you don't get anything indicating about the other's close call..... this is due to the thing that both sides of the pipe are still open (by the same process) so it will not receive anything because it can still write to the pipe.
Any other kind of misbehaviour could be explained if you had posted any code, but you didn't, so IMHO, thi answer is still incomplete, but the number of different scenarios is difficult to enumerate, so I'll be pendant of any update to your question with some faulty (or needed of explanation) code.
I'm writing a library that should execute a program in a child process, capture the output, and make the output available in a line by line (string vector) way. There is one vector for STDOUT, one for STDERR, and one for "STDCOMBINED", i.e. all output in the order it was printed by the program. The child process is connected via two pipes to a parent process. One pipe for STDOUT and one for STDERR. In the parent process I read from the read-ends of the pipes, in the child process I dup2()'ed STDOUT/STDERR to the write ends of the pipes.
My problem:
I'd like to capture STDOUT, STDERR, and "STDCOMBINED" (=both in the order they appeared). But the order in the combined vector is different to the original order.
My approach:
I iterate until both pipes show EOF and the child process exited. At each iteration I read exactly one line (or EOF) from STDOUT and exactly one line (or EOF) from STDERR. This works so far. But when I capture out the lines as they come in the parent process, the order of STDOUT and STDERR is not the same as if I execute the program in a shell and look at the output.
Why is this so and how can I fix this? Is this possible at all? I know in the child process I could redirect STDOUT and STDERR both to a single pipe but I need STDOUT and STDERR separately, and "STDCOMBINED".
PS: I'm familiar with libc/unix system calls, like dup2(), pipe(), etc. Therefore I didn't post code. My question is about the general approach and not a coding problem in a specific language. I'm doing it in Rust against the raw libc bindings.
PPS: I made a simple test program, that has a mixup of 5 stdout and 5 stderr messages. That's enough to reproduce the problem.
At each iteration I read exactly one line (or EOF) from STDOUT and exactly one line (or EOF) from STDERR.
This is the problem. This will only capture the correct order if that was exactly the order of output in the child process.
You need to capture the asynchronous nature of the beast: make your pipe endpoints nonblocking, select* on the pipes, and read whatever data is present, as soon as select returns. Then you'll capture the correct order of the output. Of course now you can't be reading "exactly one line": you'll have to read whatever data is available and no more, so that you won't block, and maintain a per-pipe buffer where you append new data, extract any lines that are present, shove the unprocessed output to the beginning, and repeat. You could also use a circular buffer to save a little bit of memcpy-ing, but that's probably not very important.
Since you're doing this in Rust, I presume there's already a good asynchronous reaction pattern that you could leverage (I'm spoiled with go, I guess, and project the hopes on the unsuspecting).
*Always prefer platform-specific higher-performance primitives like epoll on Linux, /dev/poll on Solaris, pollset &c. on AIX
Another possibility is to launch the target process with LD_PRELOAD, with a dedicated library that it takes over glibc's POSIX write, detects writes to the pipes, and encapsulates such writes (and only those) in a packet by prepending it with a header that has an (atomically updated) process-wide incrementing counter stored in it, as well as the size of the write. Such headers can be easily decoded on the other end of the pipe to reorder the writes with a higher chance of success.
I think it's not possible to strictly do what you want to do.
If you think about how it's done when running a command in an interactive shell, what happens is that both stdout and stderr point to the same file descriptor (the TTY), so the total ordering is correct by means of synchronization against the same file.
To illustrate, imagine what happens if the child process has 2 completely independent threads, one only writing to stderr, and to other only writing to stdout. The total ordering would depend on however the scheduler decided to schedule these threads, and if you wanted to capture that, you'd need to synchronize those threads against something.
And of course, something can write thousands of lines to stdout before writing anything to stderr.
There are 2 ways to relax your requirements into something workable:
Have the user pass a flag waiving separate stdout and stderr streams in favor of a correct stdcombined, and then redirect both to a single file descriptor. You might need to change the buffering settings (like stdbuf does) before you execute the process.
Assume that stdout and stderr are "reasonably interleaved", an assumption pointed out by #Nate Eldredge, in which case you can use #Unslander Monica's answer.
I tried to see what happens if I read something from keyboard while I have multiple processes with fork() (in my case there are two children and a parent) and I discovered the following problem: I need to tell the parent to wait for children's processes, otherwise the program behaves strangely.
I did a research and I found that the problem is with the parent, he needs to wait for the child's process to end because if the parent's process ends first somehow he closes the STDIN, am I right? But also I found that every process has a copy of STDIN so my question is:
Why it works this way and why only the parent has the problem with STDIN and the children not, I mean why if the child's process ends first doesn't affect STDIN but if the parent's process ends first it does affect STDIN?
Here are my tests:
I ran the program without wait() and after I typed a number the program stopped, but then I pressed enter two more times and the other two messages from printf() appeared.
When I ran the program with wait() everything worked fine, every process called scanf() separately and read a different number.
Well, a lot of stuff is going on here. I will try to explain it step by step.
When you start your terminal, the terminal creates a special file having path /dev/pts/<some number>. Then it starts your shell (which is bash in this case) and links the STDIN, STDOUT and STDERR of the bash process to this special file. This file is called a special file because it doesn't actually exist on your hard disk. Instead, whatever you write to this file, it goes directly to the terminal and the terminal renders it on the screen. (Similarly, whenever you try to read from this file, the read blocks until someone types something at the terminal).
Now when you launch your program by typing ./main, bash calls the fork function in order create a new process. The child process execs your executable file, while the parent process waits for the child to terminate. Your program then calls fork twice and we have three processes trying to read their STDINs, ie the same file /dev/pts/something. (Remember that calling fork and exec duplicates and preserves the file descriptors respectively).
The three processes are in race condition. When you enter something at the terminal, one of the three processes will receive it (99 out of 100 times it would be the parent process since the children have to do more work before reaching scanf statement).
So, parent process prints the number and exits first. The bash process that was waiting for the parent to finish, resumes and puts the STDIN into a so called "non-canonical" mode, and calls read in order to read the next command. Now again, three processes (Child1, Child2 and bash) are trying to read STDIN.
Since the children are trying to read STDIN for a longer time, the next time you enter something it will be received by one of the children, rather than bash. So you think of typing, say, 23. But oops! Just after you press the 2 key, you get Your number is: 2. You didn't even press the Enter key! That happened because of this so called "non-canonical" mode. I won't be going into what and why is that. But for now, to make things easier, use can run your program on sh instead of bash, since sh doesn't put STDIN into non-canonical mode. That will make the picture clear.
TL;DR
No, parent process closing its STDIN doesn't mean that its children or other process won't be able to use it.
The strange behavior you are seeing is because when the parent exits, bash puts the pty (pseudo terminal) into non-canonical mode. If you use sh instead, you won't see that behavior. Read up on pseudo terminals, and line discipline if you want to have a clear understading.
The shell process will resume as soon as the parent exits.
If you use wait to ensure that parents exits last, you won't have any problem, since the shell won't be able to run along with your program.
Normally, bash makes sure that no two foreground processes read from STDIN simultaneously, so you don't see this strange behavior. It does this by either piping STDOUT of one program to another, or by making one process a background process.
Trivia: When a background process tries to read from its STDIN, it is sent a signal SIGTTIN, which stops the process. Though, that's not really relevant to this scenario.
There are several issues that can happen when multiple processes try to do I/O to the same TTY. Without code, we can't tell which may be happening.
Trying to do I/O from a background process group may deliver a signal: SIGTTIN for input (usually enabled), or SIGTTOU for output (usually disabled)
Buffering: if you do any I/O before the fork, any data that has been buffered will be there for both processes. Under some conditions, using fflush may help, but it's better to avoid buffering entirely. Remember that, unlike output buffering, it is impossible to buffer input on a line-by-line basis (although you can only buffer what is available, so it may appear to be line-buffered at first).
Race conditions: if more than one process is trying to read the same pipe-like file, it is undefined which one will "win" and actually get the input each time it is available.
So, I was given this one line script:
echo test | cat | grep test
Could you please explain to me how exactly that would work given the following system calls: pipe(), fork(), exec() and dup2()?
I am looking for an general overview here and mainly the sequence of operations.
What I know so far is that the shell will fork using fork() and the script's code will replace the shell's one by using the exec(). But what about pipe and dup2? How do they fall in place?
Thanks in advance.
First consider a simpler example, such as:
echo test | cat
What we want is to execute echo in a separate process, arranging for its standard output to be diverted into the standard input of the process executing cat. Ideally this diversion, once setup, would require no further intervention by the shell — the shell would just calmly wait for both processes to exit.
The mechanism to achieve that is called the "pipe". It is an interprocess communication device implemented in the kernel and exported to the user-space. Once created by a Unix program, a pipe has the appearance of a pair of file descriptors with the peculiar property that, if you write into one of them, you can read the same data from the other. This is not very useful within the same process, but keep in mind that file descriptors, including but not limited to pipes, are inherited across fork() and even accross exec(). This makes pipe an easy to set up and reasonably efficient IPC mechanism.
The shell creates the pipe, and now owns a set of file descriptors belonging to the pipe, one for reading and one for writing. These file descriptors are inherited by both forked subprocesses. Now only if echo were writing to the pipe's write-end descriptor instead of to its actual standard output, and if cat were reading from the pipe's read-end descriptor instead of from its standard input, everything would work. But they don't, and this is where dup2 comes into play.
dup2 duplicates a file descriptor as another file descriptor, automatically closing the new descriptor beforehand. For example, dup2(1, 15) will close file descriptor 1 (by convention used for the standard output), and reopen it as a copy of file descriptor 15 — meaning that writing to the standard output will in fact be equivalent to writing to file descriptor 15. The same applies to reading: dup2(0, 8) will make reading from file descriptor 0 (the standard input) equivalent to reading from file descriptor 8. If we proceed to close the original file descriptor, the open file (or a pipe) will have been effectively moved from the original descriptor to the new one, much like sci-fi teleports that work by first duplicating a piece of matter at a remote location and then disintegrating the original.
If you're still following the theory, the order of operations performed by the shell should now be clear:
The shell creates a pipe and then fork two processes, both of which will inherit the pipe file descriptors, r and w.
In the subprocess about to execute echo, the shell calls dup2(1, w); close(w) before exec in order to redirect the standard output to the write end of the pipe.
In the subprocess about to execute cat, the shell calls dup2(0, r); close(r) in order to redirect the standard input to the read end of the pipe.
After forking, the main shell process must itself close both ends of the pipe. One reason is to free up resources associated with the pipe once subprocesses exit. The other is to allow cat to actually terminate — a pipe's reader will receive EOF only after all copies of the write end of the pipe are closed. In steps above, we did close the child's redundant copy of the write end, the file descriptor 15, right after its duplication to 1. But the file descriptor 15 must also exist in the parent, because it was inherited under that number, and can only be closed by the parent. Failing to do that leaves cat's standard input never reporting EOF, and its cat process hanging as a consequence.
This mechanism is easily generalized it to three or more processes connected by pipes. In case of three processes, the pipes need to arrange that echo's output writes to cat's input, and cat's output writes to grep's input. This requires two calls to pipe(), three calls to fork(), four calls to dup2() and close (one for echo and grep and two for cat), three calls to exec(), and four additional calls to close() (two for each pipe).
I'm trying to find out if a child process is waiting for user input (without parsing its output). Is it possible, in C on Unix, to determine if a pipe's read end currently has a read() call blocking?
The thing is, I have no control over the programs exec'd in the child processes. They print all kinds of verbose garbage which I would usually want to redirect to /dev/null. Occasionally though one will prompt the user for something. (With the prompt having no reliable format.) So my idea was:
In a loop:
Drain child's stdout, append it to a temporary buffer.
Check (no idea how) if the child is asking for user input, in which case the buffer is printed to stdout.
When the child exits, throw away the buffer.
The thing is, I have no control over the programs exec'd in the child processes. They print all kinds of verbose garbage which I would usually want to redirect to /dev/null. Occasionally though one will prompt the user for something. (With the prompt having no reliable format.) So my idea was:
In a loop:
Drain child's stdout, append it to a temporary buffer.
Check (no idea how) if the child is asking for user input, in which case the buffer is printed to stdout.
When the child exits, throw away the buffer.
You have these options:
if you know that the child will need certain input (such as shell that will read a command), just write to a pipe
if you assume the child won't read anything usually, but may do it sometimes, you probably need something like job control in the shell (use a terminal for communication with the child, use process groups and TIOCSPGRP ioctl on the terminal to get the child to the background; the child will get SIGTTIN when it tries to read from the terminal, and you can wait() for that). This is how bash handles things like "(sleep 10; read a;)&"
if you don't know what to write, or you have more possibilities, you will have to parse the output
That sounds as if you were trying to supervise dpkg where occasionally some post-inst script queries the admin whether it may override some config file.
Anyway, you may want to look at how strace works:
strace -f -etrace=read your.program
Of course you need to keep track of which fds are the pipes you write about, but you probably need only stdin, anyway.
I don't think that's true: For example, right before calling read() on the reader side, the pipe would have a reader that isn't actually reading.
You would typically just write to the pipe, or use select or poll. If you need a handshake mechanism you can do that out of band various ways or come up with and in-band protocol.
I don't know if there is a built-in way to know if a reader on the other end is blocking. Why do you need to know this?
If I recall correctly, you can not have a pipe with no reader which means that you have either a read(2) or a select(2) syscal pending at all time.