I have a quite unusual problem in one of my bash scripts. I want to do something (in fact I want to create/remve LVs, this is MWE) like this:
#! /bin/bash
# ...
exec {flock}>/tmp/lock
# Do something with fd ${flock} e.g.
flock -n ${flock} || exit 1
# ...
lvs ${flock}>&-
# ...
The problem is the ${flock}>&-. Why do I want this? The LVM tools complain with a warning about any opened file descriptors except for stdin, stdout and stderr. So when I drop this small redirecting part, the script works but writes out a warning message.
Thus I wanted to redirect the fd $flock only for the LVM command to be closed. I do not want to close the file but only redirect for this single command invocation.
In my case $flock is set to 10 (first free fd greater or equal to 10, see man bash). However I do not get the corresponding fd remapped as sketched above. Instead the 10 is considered a parameter of the (lvs) command and the stdout should be redirected. Of course this is not what I intend.
If I hardcode 10>&- this works but is very bad style. For now I switched to completely hardcode the fd in the whole file. Nevertheless I would like to know how it would be done correctly.
Don't use the dollar
Your example code has:
lvs ${flock}>&-
The correct syntax is:
lvs {flock}>&-
From the redirection section of the Bash manual:
Each redirection that may be preceded by a file descriptor number may instead be preceded by a word of the form {varname}. In this case, for each redirection operator except >&- and <&-, the shell will allocate a file descriptor greater than 10 and assign it to {varname}. If >&- or <&- is preceded by {varname}, the value of varname defines the file descriptor to close. If {varname} is supplied, the redirection persists beyond the scope of the command, allowing the shell programmer to manage the file descriptor himself.
(emphasis mine)
I have seen this question:
Shell redirection i/o order.
But I have another question. If this line fails to redirect stderr to file:
ls -xy 2>&1 1>file
Then why this line can redirect stderr to grep?
ls -xy 2>&1 | grep ls
I want to know how it is actually being run underneath.
It is said that 2>&1 redirects stderr to a copy of stdout. What does "a copy of stdout" mean? What is actually being copied?
The terminal registers itself (through the OS) for sending and receiving through the standard streams of the processes it creates, right? Does the other redirections go through the OS as well (I don't think so, as the terminal can handle this itself)?
The pipe redirection (connecting standard output of one command to the stdin of the next) happens before the redirection performed by the command.
That means by the time 2>&1 happens, the stdout of ls is already setup to connect to stdin of grep.
See the man page of bash:
Pipelines
The standard output of command is connected via a pipe to
thestandard input of command2. This connection is performed before
anyredirections specified by the command (see REDIRECTION below). If
|&is used, command's standard error, in addition to its
standardoutput, is connected to command2's standard input through the
pipe;it is shorthand for 2>&1 |. This implicit redirection of
thestandard error to the standard output is performed after
anyredirections specified by the command.
(emphasis mine).
Whereas in the former case (ls -xy 2>&1 1>file), nothing like that happens i.e. when 2>&1 is performed the stdout of ls is still connected to the terminal (and hasn't yet been redirected to the file).
That answers my first question. What about the others?
Well, your second question has already been answered in the comments. (What is being duplicated is a file descriptor).
As to your last question(s),
The terminal registers itself (through the OS) for sending and receiving through the standard streams of the processes it creates, right? Does the other redirections go through the OS as well (I don't think so, as the terminal can handle this itself)?
it is the shell which attaches the standard streams of the processes it creates (pipes first, then <>’s, as you have just learned). In the default case, it attaches them to its own streams, which might be attached to a tty, with which you can interact in a number of ways, usually a terminal emulation window, or a serial console, whatever. Terminal is a very ambiguous word.
I would like to know if it's possible to get the output redirection file name as a parameter in bash?
For example :
./myscript.sh parameter1 > outputfile
Is there a way to get "outputfile" as a parameter like $2? In my script I have to do few operations in outputfile but I don't know which file I have to update... The second problem is, this script is already running and used by several tasks so I cannot change the user input...
Best regards
Redirections are not parameters to the program. When a program's output is redirected, the shell opens the file and connects file descriptor 2 to it before running the program. The program then simply writes to fd 2 (aka stdout) and it goes to the file.
On Linux and similar systems you can use /dev/stdout, which is a symbolic link to the process's stdout file.
How to explain the output of cat /etc/passwd | cat </etc/issue?
In this case, the second cat receives contents from /etc/passwd as $STDIN and again /etc/issue is redirected. Why there is only /etc/issue left?
What's more, cat </etc/passwd </etc/issue only outputs the contents in /etc/issue. Is /etc/passwd overwritten?
I am not looking for a solution how to cat two files, but confused with how pipeline works.
Piping and redirection are processed from left to right.
So first the input of cat is redirected to the pipe. Then it is redirected to /etc/issue. Then the program is run, using the last redirection, which is the file.
When you do cat <file1 <file2, stdin is first redirected to file1, then it is redirected to file2. Then the program is run, and it gets its input from the last redirection.
It's like variable assignments. If you do:
stdin=passwd
stdin=issue
The value of stdin at the end is the last one assigned.
This is explained in the bash documentation, in the first paragraph of the section on Redirection:
Before a command is executed, its input and output may be redirected using a special notation interpreted by the shell. Redirection may also be used to open and close files for the current shell execution environment. The following redirection operators may precede or appear anywhere within a simple command or may follow a command. Redirections are processed in the order they appear, from left to right.
(emphasis mine). I assume it's also in the POSIX shell specification, I haven't bothered to look it up. This is how Unix shells have always behaved.
The pipe is created first: the standard output of cat /etc/passwd is sent to write side of the pipe, and the standard input of cat </etc/issue is set to the read side of the pipe. Then the command on each half of the pipe is processed. There's no other I/O redirection on the LHS, but on the RHS, the standard input is redirected so it comes from /etc/issue. That means there's nothing actually reading the read end of the pipe, so the LHS cat is terminated with a SIGPIPE (probably; alternatively, it writes data to the pipe but no process ever reads it). The LHS cat never knows about the pipe input — it only has the the file input for its standard input.
I often use pipes in Bash, e.g.:
dmesg | less
Although I know what this outputs, it takes dmesg and lets me scroll through it with less, I do not understand what the | is doing. Is it simply the opposite of >?
Is there a simple, or metaphorical explanation for what | does?
What goes on when several pipes are used in a single line?
Is the behavior of pipes consistent everywhere it appears in a Bash script?
A Unix pipe connects the STDOUT (standard output) file descriptor of the first process to the STDIN (standard input) of the second. What happens then is that when the first process writes to its STDOUT, that output can be immediately read (from STDIN) by the second process.
Using multiple pipes is no different than using a single pipe. Each pipe is independent, and simply links the STDOUT and STDIN of the adjacent processes.
Your third question is a little bit ambiguous. Yes, pipes, as such, are consistent everywhere in a bash script. However, the pipe character | can represent different things. Double pipe (||), represents the "or" operator, for example.
In Linux (and Unix in general) each process has three default file descriptors:
fd #0 Represents the standard input of the process
fd #1 Represents the standard output of the process
fd #2 Represents the standard error output of the process
Normally, when you run a simple program these file descriptors by default are configured as following:
default input is read from the keyboard
Standard output is configured to be the monitor
Standard error is configured to be the monitor also
Bash provides several operators to change this behavior (take a look to the >, >> and < operators for example). Thus, you can redirect the output to something other than the standard output or read your input from other stream different than the keyboard. Specially interesting the case when two programs are collaborating in such way that one uses the output of the other as its input. To make this collaboration easy Bash provides the pipe operator |. Please note the usage of collaboration instead of chaining. I avoided the usage of this term since in fact a pipe is not sequential. A normal command line with pipes has the following aspect:
> program_1 | program_2 | ... | program_n
The above command line is a little bit misleading: user could think that program_2 gets its input once the program_1 has finished its execution, which is not correct. In fact, what bash does is to launch ALL the programs in parallel and it configures the inputs outputs accordingly so every program gets its input from the previous one and delivers its output to the next one (in the command line established order).
Following is a simple example from Creating pipe in C of creating a pipe between a parent and child process. The important part is the call to the pipe() and how the parent closes fd1 (writing side) and how the child closes fd1 (writing side). Please, note that the pipe is a unidirectional communication channel. Thus, data can only flow in one direction: fd1 towards fd[0]. For more information take a look to the manual page of pipe().
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
int main(void)
{
int fd[2], nbytes;
pid_t childpid;
char string[] = "Hello, world!\n";
char readbuffer[80];
pipe(fd);
if((childpid = fork()) == -1)
{
perror("fork");
exit(1);
}
if(childpid == 0)
{
/* Child process closes up input side of pipe */
close(fd[0]);
/* Send "string" through the output side of pipe */
write(fd[1], string, (strlen(string)+1));
exit(0);
}
else
{
/* Parent process closes up output side of pipe */
close(fd[1]);
/* Read in a string from the pipe */
nbytes = read(fd[0], readbuffer, sizeof(readbuffer));
printf("Received string: %s", readbuffer);
}
return(0);
}
Last but not least, when you have a command line in the form:
> program_1 | program_2 | program_3
The return code of the whole line is set to the last command. In this case program_3. If you would like to get an intermediate return code you have to set the pipefail or get it from the PIPESTATUS.
Every standard process in Unix has at least three file descriptors, which are sort of like interfaces:
Standard output, which is the place where the process prints its data (most of the time the console, that is, your screen or terminal).
Standard input, which is the place it gets its data from (most of the time it may be something akin to your keyboard).
Standard error, which is the place where errors and sometimes other out-of-band data goes. It's not interesting right now because pipes don't normally deal with it.
The pipe connects the standard output of the process to the left to the standard input of the process of the right. You can think of it as a dedicated program that takes care of copying everything that one program prints, and feeding it to the next program (the one after the pipe symbol). It's not exactly that, but it's an adequate enough analogy.
Each pipe operates on exactly two things: the standard output coming from its left and the input stream expected at its right. Each of those could be attached to a single process or another bit of the pipeline, which is the case in a multi-pipe command line. But that's not relevant to the actual operation of the pipe; each pipe does its own.
The redirection operator (>) does something related, but simpler: by default it sends the standard output of a process directly to a file. As you can see it's not the opposite of a pipe, but actually complementary. The opposite of > is unsurprisingly <, which takes the content of a file and sends it to the standard input of a process (think of it as a program that reads a file byte by byte and types it in a process for you).
In short, as described, there are three key 'special' file descriptors to be aware of. The shell by default send the keyboard to stdin and sends stdout and stderr to the screen:
A pipeline is just a shell convenience which attaches the stdout of one process directly to the stdin of the next:
There are a lot of subtleties to how this works, for example, the stderr stream might not be piped as you would expect, as shown below:
I have spent quite some time trying to write a detailed but beginner friendly explanation of pipelines in Bash. The full content is at:
https://effective-shell.com/docs/part-2-core-skills/7-thinking-in-pipelines/
A pipe takes the output of a process, by output I mean the standard output (stdout on UNIX) and passes it on the standard input (stdin) of another process. It is not the opposite of the simple right redirection > which purpose is to redirect an output to another output.
For example, take the echo command on Linux which is simply printing a string passed in parameter on the standard output. If you use a simple redirect like :
echo "Hello world" > helloworld.txt
the shell will redirect the normal output initially intended to be on stdout and print it directly into the file helloworld.txt.
Now, take this example which involves the pipe :
ls -l | grep helloworld.txt
The standard output of the ls command will be outputed at the entry of grep, so how does this work?
Programs such as grep when they're being used without any arguments are simply reading and waiting for something to be passed on their standard input (stdin). When they catch something, like the ouput of the ls command, grep acts normally by finding an occurence of what you're searching for.
Pipes are very simple like this.
You have the output of one command. You can provide this output as the input into another command using pipe. You can pipe as many commands as you want.
ex:
ls | grep my | grep files
This first lists the files in the working directory. This output is checked by the grep command for the word "my". The output of this is now into the second grep command which finally searches for the word "files". Thats it.
The pipe operator takes the output of the first command, and 'pipes' it to the second one by connecting stdin and stdout.
In your example, instead of the output of dmesg command going to stdout (and throwing it out on the console), it is going right into your next command.
| puts the STDOUT of the command at left side to the STDIN of the command of right side.
If you use multiple pipes, it's just a chain of pipes. First commands output is set to second commands input. Second commands output is set to next commands input. An so on.
It's available in all Linux/widows based command interpreter.
All of these answere are great. Something that I would just like to mention, is that a pipe in bash (which has the same concept as a unix/linux, or windows named pipe) is just like a pipe in real life.
If you think of the program before the pipe as a source of water, the pipe as a water pipe, and the program after the pipe as something that uses the water (with the program output as water), then you pretty much understand how pipes work.
And remember that all apps in a pipeline run in parallel.
Regarding the efficiency issue of pipe:
A command can access and process the data at its input before previous pipe command to complete that means computing power utilization efficiency if resources available.
Pipe does not require to save output of a command to a file before next command to access its input ( there is no I/O operation between two commands) that means reduction in costly I/O operations and disk space efficiency.
If you treat each unix command as a standalone module,
but you need them to talk to each other using text as a consistent interface,
how can it be done?
cmd input output
echo "foobar" string "foobar"
cat "somefile.txt" file *string inside the file*
grep "pattern" "a.txt" pattern, input file *matched string*
You can say | is a metaphor for passing the baton in a relay marathon.
Its even shaped like one!
cat -> echo -> less -> awk -> perl is analogous to cat | echo | less | awk | perl.
cat "somefile.txt" | echo
cat pass its output for echo to use.
What happens when there is more than one input?
cat "somefile.txt" | grep "pattern"
There is an implicit rule that says "pass it as input file rather than pattern" for grep.
You will slowly develop the eye for knowing which parameter is which by experience.