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From: charles.deledalle@gmail.com
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Subject: $LINENO in a bash script using subshells in if statements
Date: Fri,  5 Jul 2019 15:30:20 -0700 (PDT)
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Configuration Information [Automatically generated, do not change]:
Machine: x86_64
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS: -g -O2 -fdebug-prefix-map=/build/bash-LQgi2O/bash-5.0=. -fstack-protector-strong -Wformat -Werror=format-security -Wall -Wno-parentheses -Wno-format-security
uname output: Linux denver 5.0.0-16-generic #17-Ubuntu SMP Wed May 15 10:52:21 UTC 2019 x86_64 x86_64 x86_64 GNU/Linux
Machine Type: x86_64-pc-linux-gnu

Bash Version: 5.0
Patch Level: 3
Release Status: release

Description:
	Using a subshell environment in an if statement screws up my $LINENO variable, see minimal reproducible code below.
	The problem occurs only if the condition is true (meaning the subshell is executed). Same problem occurs if the subshell is in a while/for loop iterating at least once.
	The problem occurs with Bash 5.0.3. It does not occur with 4.2.3 or 3.2.57. See discussion here: https://stackoverflow.com/questions/56909685/weird-behavior-of-lineno-in-a-bash-script-using-subshells-in-if-statements

Repeat-By:
	Here is the minimal reproducible code

	1. #!/bin/bash
	2. if true ; then
	3.     (echo dummy)
	4. fi
	5. echo "Line no:" $LINENO

	Shows "Line no: 4" instead of 5.

Fix:
	Problem can be fixed as

	1. #!/bin/bash
	2. (if true ; then
	3.     (echo dummy)
	4. fi)
	5. echo "Line no:" $LINENO

	but this is quite a hugly workaround.