So far, I know the piping mechanism as a way to connect a series of commands by connecting the stdout of one command to the stdin of the next command until the last command is reached, which connects its stdout with the display or a file.

Would it be possible, however, to make a loop out of commands, so the last command's stdout connects to the first command's stdin and maybe by using tee somehow the changing values of a certain output could be displayed?

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    I edited your post with a personal opinion; feel free to revert it, of course, but I'm just trying to keep the question clear and ... non-revolting. – Jeff Schaller May 27 '17 at 20:13
  • if I made a calvinistic protestant happy by improving the world by making it more like god imagined it with the power of the holy editor function, I am more than pleased – sharkant May 27 '17 at 20:18
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    Do you have a use case for where something like this would be a solution? Or is it a purely intellectual exercise? – Kusalananda May 27 '17 at 20:33
  • i am bloody beginner, thus I do not know enough for making reasonable solutions for anything, so like chanting A B C i am bound to intellectual exercises and speculations, hoping not asking too stupid questions or becomming notorious for contributing waste without benefit – sharkant May 27 '17 at 20:39
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    @JeffSchaller I was like "what could Jeff possibly mean, revolting" and then i clicked the edit history – cat May 28 '17 at 1:57

Well you certainly could by just make a loop and use variables:

while true; do
    a=$(echo "$a" | grep "Hey" | cut -d" " -f2 | tee -a log)

That would save the last output which would be used at the beginning again

  • a = command1|command2|command3 so this structure would save the std output of command3 into variable a, but what if command3 does not have one but a series of values? – sharkant May 27 '17 at 20:49
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    It would all be saved as one whole string in the Variable a and be used again in command 1 – ADDB May 27 '17 at 20:51
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    I'm not sure if this answers the actual question: that's a whole different kind of a loop. – ilkkachu May 28 '17 at 8:30
  • Nah it doesn't, he accepted it just for fun. He asked how do built a loop that's using the stdout as stdin, and that is what this loop does – ADDB May 28 '17 at 8:32

I'm not sure about all the shells there are, but in Bash it's possible, though not with unnamed pipes. So not with the | symbol. But if you create a named pipe:

mkfifo fifo

Then you can use it:

<fifo cat | cat >fifo &

Now the pipeline works in the background, but does nothing. But if you feed the pipe from outside the pipeline:

echo x >fifo

The pipeline will unblock and go on forever. Or until you drain the pipe:

cat fifo

The output will appear once:


To make this a little sophisticated, the pipeline may be this:

<fifo cat | xargs -I@ echo @x >fifo &

So it will add an x to the output at each iteration. Of course it will, but only once the iterations start, meaning as soon as the pipe unblock, that is as soon as there's something to read. As previously, this can be started by hand:

echo x >fifo

And now take a look at what top shows. There should be quite a lot of activity of both cat and xargs.

And the same as before, if you drain the pipeline, you should see a lot of xs in the terminal, and the pipeline will block.

It would be a valid question, why does the pipeline get drained. Why is the cat command committed in the terminal leaving nothing in the circuit. I don't know this.

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    This is heavily dependent on the size of the standard IO buffer size. If any data between two parts of the pipeline exceeds the buffer size, the pipeline would block. – Kusalananda May 27 '17 at 20:53
  • @Kusalananda That will be the limit. And it may be an answer why the pipeline gets drained. – Tomasz May 27 '17 at 20:55

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