[SOLVED] is this a bug in bash?

pan64 · 07-03-2019, 02:16 AM

or how should it really work?
https://github.com/koalaman/shellcheck/issues/1585

based on my knowledge it looks like a bug. What do you think about it?

NevemTeve · 07-03-2019, 03:39 AM

You mean a bug in shell-check? No, it warns you because the redirection makes the cycle run in a sub-shell, and you canot set the parent shell's variable from the sub-shell.

MadeInGermany · 07-03-2019, 03:46 AM

The pipe forces the loop block to run in a sub shell.
Variables and settings in a sub shell are not copied back to the main shell.
All shells (bash,ksh,zsh,dash,mksh) agree in that.

More precise: the code parts left and right from a pipe are forced into a sub shell.

The only exception is ksh: it runs the last part in the main shell:

Code:

failed=0
echo "hello" |
while read line; do
  echo "FAILED"
  failed=1
done
echo "Failed: $failed"

Only ksh says:

Code:

Failed: 1

BTW this proves that mksh, despite its name, is more a Posix shell than a ksh.

pan64 · 07-03-2019, 04:06 AM

looks like you misunderstood.
The original example does not contain pipe before the while loop.

Code:

#!/bin/bash

echo $BASHPID
failed=0
while read -r _; do
  echo "FAILED"
  echo $BASHPID
  failed=1
done < <(echo hello) | sort -u
echo "Failed: $failed"

And it looks like the while loop still runs in another shell

MadeInGermany · 07-03-2019, 04:37 AM

The original example has the loop block left from the pipe. That is always in a sub shell.
I tried to say that in my first sentence.

pan64 · 07-03-2019, 06:01 AM

Yes you are right, I know about that. But this is a bit different case.

crts · 07-03-2019, 07:58 AM

Hello OP,
Since you have not marked the thread as solved I am not sure if your question is answered to your satisfaction. If you are still looking for an explanation then here is the excerpt from the manpage that explains the behaviour:

Code:

   Pipelines
       A  pipeline  is  a sequence of one or more commands separated by one of the control operators | or |&.
       The format for a pipeline is:

              [time [-p]] [ ! ] command [ [|⎪|&] command2 ... ]

       ...

       Each command in a pipeline is executed as a separate process (i.e., in a subshell).

If you are looking for a workaround then process substitution might help:

Code:

#!/usr/bin/bash

failed=0

# process substitution sets failed correctly and sorts
while read -r l; do
        echo $l
        failed=1
done < <(echo "world";echo "hello") > >(sort -u)
echo "Failed: $failed"

Keep in mind though, that 'sort' may output to the screen after the last 'echo'.
Of course, you could also use temporary files to store the value.

pan64 · 07-03-2019, 08:19 AM

Ok, let's have a simple example:

Code:

user@host:/tmp$ echo $$
1984
user@host:/tmp$ echo $$ | sort
1984

Code:

user@host:/tmp$ echo $BASHPID
1984
user@host:/tmp$ echo $BASHPID | sort
27658

based on this (and the man page of bash) this echo is running in a subshell which does not require bash to be reinitialized, and the PIDs ($$) are the same.

something learned again