Bash: Quotes getting stripped when a command is passed as argument to a function

I am trying to implement a dry run kind of mechanism for my script and facing the issue of quotes getting stripped off when a command is passed as an argument to a function and resulting in unexpected behavior.

dry_run () {
    echo "$@"
    #printf '%q ' "$@"

    if [ "$DRY_RUN" ]; then
        return 0
    fi

    "$@"
}


email_admin() {
    echo " Emailing admin"
    dry_run su - $target_username  -c "cd $GIT_WORK_TREE && git log -1 -p|mail -s '$mail_subject' $admin_email"
    echo " Emailed"
    }

Output is:

su - webuser1 -c cd /home/webuser1/public_html && git log -1 -p|mail -s 'Git deployment on webuser1' [email protected]

Expected:

su - webuser1 -c "cd /home/webuser1/public_html && git log -1 -p|mail -s 'Git deployment on webuser1' [email protected]"

With printf enabled instead of echo:

su - webuser1 -c cd\ /home/webuser1/public_html\ \&\&\ git\ log\ -1\ -p\|mail\ -s\ \'Git\ deployment\ on\ webuser1\'\ [email protected]

Result:

su: invalid option -- 1

That shouldn't be the case if quotes remained where they were inserted. I have also tried using "eval", not much difference. If i remove the dry_run call in email_admin and then run script, it work great.

Try using \" instead of just ".

"$@" should work. In fact it works for me in this simple test case:

dry_run()
{
    "$@"
}

email_admin()
{
    dry_run su - foo -c "cd /var/tmp && ls -1"
}

email_admin

Output:

./foo.sh 
a
b

Edited to add: the output of echo $@ is correct. The " is a meta-character and not part of the parameter. You can prove that it is correctly working by adding echo $5 to dry_run(). It will output everything after -c