I thought that when i running below commands
sleep 10 | sleep 2 | sleep 5
Linux process will be
86014 ttys002 0:00.03 bash 86146 ttys002 0:00.00 sleep 10
when sleep 10 is end -> sleep 2 (when sleep 2 is end)-> sleep 5
that’s i thought
But in Linux bash sleep 10, sleep 2, sleep 5 are in ps same time
Standard output of sleep 10 process will be redirected to sleep 5‘s process
But, in that case, sleep 5‘s process will be finished before sleep 10
I’m confused, any google keywords or concepts of that phenomenon?
(I’m not good at English, maybe it’s hard to understand this text🥲. Thank you)
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Answer
I think you expect the commands to be run in sequence. But that is not what a pipe does.
To run two commands in sequence you use the ;, what it is called a command list, I think:
$ time ( sleep 1 ; sleep 2 ) real 0m3.004s
You can also do command lists with && (or ||) so that the sequence is interrupted if one command returns failure (or success).
But when you run two commands with | both are run in parallel, and the stdout of the first is connected to the stdinof the second. That way, the pipe acts as a synchronization object:
- If the second command is faster and empties the pipe, when it reads it will wait for more data
- If the first command is faster and writes too much data to the pipe, its buffer will fill up and it will block until some data is read.
- Additionally, if the second command dies, as soon as the first one writes to
stdoutit will get aSIGPIPE, and possibly die.
(Note what would happen if your programs were not run concurrently: your first program could write megabytes of text to stdout, and with nobody to read it, the pipe would overflow!)
But since sleep does not read or write to the console, when you do sleep 1 | sleep 2 nothing special happens and both are run concurrently.
The same happens with 3 or any other number of commands in your pipe.
The net effect is that the full sleep is the longest:
$ time ( sleep 1 | sleep 2 | sleep 3 | sleep 4 ) real 0m4.004s