Line continuation + line comment, on the same line? [duplicate]

Question

I have a call with a pretty long list of arguments in my shell script:

foo-command \
    --important-argument $ONE \
    --indispensable-flag $ENABLED \
    --an-optional-toggle "will probably be commented out some day" \
    $ARGUMENTS \
    $MORE_ARGUMENTS

Can't I insert comments in this multiline call, can I?

For example, lets say, how can I comment out the --an-optional-toggle line?
How can I add an # in sorted order comment after $ARGUMENTS?

No matter what I try, the octothorp # symbol shadows line continuation backslash \, or vice versa. Any advise?

Answer 1

Try the backtick comment hack as proposed in an earlier response to the same question.

In your case, this would be:

foo-command \
    --important-argument $ONE \
    --indispensable-flag $ENABLED \
    `#--an-optional-toggle "will probably be commented out some day"` \
    $ARGUMENTS \
    $MORE_ARGUMENTS

Answer 2

No, but you can store the arguments in an array instead. This allows you to both comment out an individual line as well as include interspersed comments.

args=( --important-argument "$ONE"
       --indispensable-flag "$ENABLED"
       # --an-optional-toggle "will probably be commented out some day"
       $ARGUMENTS  # in sorted order
       $MORE_ARGUMENTS
     )
foo-command "${args[@]}"

Note that you will almost certainly want to make ARGUMENTS and MORE_ARGUMENTS arrays as well, so args would end up looking like

args=( --important-argument "$ONE"
       --indispensable-flag "$ENABLED"
       # --an-optional-toggle "will probably be commented out some day"
       "${ARGUMENTS[@]}"  # in sorted order
       "${MORE_ARGUMENTS[@]}"
     )

Answer 3

foo-command $(
    # hello
    printf --important-argument $ONE # free as a bird
    printf --indispensable-flag $ENABLED
    # to comment anywhere we wish
    printf --an-optional-toggle "will probably be commented out some day"
    printf $ARGUMENTS
    printf $MORE_ARGUMENTS
)

It's not perfect: echoing -n is hard because echo interprets it; quotes may be disappeared when you would prefer to keep them, etc. In fact as commenters say below, the quoted string will get mangled; maybe you can work around that, but other answers here are better if you have Bash.