When exactly it is possible to modify BASH_LINENO variable?

[Adam Ryczkowski]

I am trying to write a function that logs execution of the *next* line. 
The usage would be:

    #/bin/bash
    log=/tmp/mylog.log
    var1="some variable"

    log
    echo "Unfortunately this line gets executed twice" | tee -a 
/tmp/temp/bla-bla.tmp

The problem is that I can't reliably get to modify `BASH_LINENO[0]`. I 
swear, that I used to have success in it, but now everytime I change its 
value (both inside or outside the function body) the assignment gets 
ignored. Is there any way to get the bash to skip execution of the next 
line?

The script goes along these lines:

    function log()
    {
        case $- in *x*) USE_X="-x";; *) USE_X=;; esac
        set +x
        if [ -n "$log" ]; then
        file=${BASH_SOURCE[1]##*/}
        linenr=$((${BASH_LINENO[0]} + 1 ))
        line=`sed "1,$((${linenr}-1)) d;${linenr} s/^ *//; q" $file`
        if [ -f /tmp/tmp.txt ]; then
            rm /tmp/tmp.txt
        fi
        echo "$line" > /tmp/tmp2.txt
        BASH_LINENO[0]=$((${BASH_LINENO[0]}+1))  #EXECUTES FINE, BUT 
DOESN'T CHANGE THE CALL STACK
        mymsg=`msg2`
        exec 3>&1 4>&2 >>/tmp/tmp.txt 2>&1
        set -x
        source /tmp/tmp2.txt
        exitstatus=$?
        set +x
        exec 1>&3 2>&4 4>&- 3>&-
        #...
        #Here goes part, that parses the /tmp/tmp.txt and appends stuff 
into $log
        #...
        fi
        if [ -n "$USE_X" ]; then
        set -x
        fi
        if [ "$exitstatus" -ne "0" ]; then
        exit $exitstatus
        fi
    }


----

This message is a crosspost from 
http://stackoverflow.com/questions/27259418/how-to-modify-call-stack-in-bash


Adam