$LINENO in a bash script using subshells in if statements

[charles . deledalle]

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.