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Bug 1241024 - case in a for loop inside subshell causes syntax error
Summary: case in a for loop inside subshell causes syntax error
Keywords:
Status: CLOSED DUPLICATE of bug 1240994
Alias: None
Product: Red Hat Enterprise Linux 6
Classification: Red Hat
Component: bash
Version: 6.6
Hardware: x86_64
OS: Linux
medium
medium
Target Milestone: rc
: ---
Assignee: Ondrej Oprala
QA Contact: BaseOS QE - Apps
URL:
Whiteboard:
Depends On:
Blocks:
TreeView+ depends on / blocked
 
Reported: 2015-07-08 10:18 UTC by Martin Kyral
Modified: 2015-08-18 09:05 UTC (History)
5 users (show)

Fixed In Version:
Doc Type: Bug Fix
Doc Text:
Clone Of: 1212775
Environment:
Last Closed: 2015-08-18 09:05:17 UTC
Target Upstream Version:
Embargoed:


Attachments (Terms of Use)

Description Martin Kyral 2015-07-08 10:18:41 UTC
mksh in RHEL 6 has the same issue. Additionally, this code:
x=$(case $i in test) echo test;; esac) which works in bash, doesn't work in mksh

Version-Release number of selected component (if applicable):
mksh-39-9.el6

+++ This bug was initially created as a clone of Bug #1212775 +++

Description of problem:

# this works
$ x=$(for i in test; do echo $i; done)
$ echo $x
test

# this too, works
$ i=test
$ x=$(case $i in test) echo test;; esac)
$ echo $x
test

# this does not
$ x=$(for i in test; do case $i in test) echo test;; esac; done)
-bash: syntax error near unexpected token `;;'

# however removing the output-into-variable makes it work
$ (for i in test; do case $i in test) echo test;; esac; done)
test


Version-Release number of selected component (if applicable):
bash-4.2.46-12.el7

How reproducible:
always

Actual results:
expression fails on syntax error

Expected results:
the subshell output is assigned to a variable

Additional info:
This *works* on my bash-4.3.30-2.fc21, so there might have been an upstream fix.

--- Additional comment from Ondrej Oprala on 2015-04-18 09:36:45 EDT ---

Hi Jiri,
Yes, this was fixed in 4.3. I have yet to look at the size of the change.

Meanwhile, the more verbose way of writing a case statement still seems to work:
x=$(for i in test; do case $i in (test) echo test;; esac; done)
#

Comment 2 Ondrej Oprala 2015-08-18 09:05:17 UTC

*** This bug has been marked as a duplicate of bug 1240994 ***


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