Concise way to output to file if argument is given, to stdout otherwise

Question

I'd like my script to output to stdout unless it gets a filename as an argument. An obvious way is like so:

if [ -e "$1" ]; then
  command_with_output >$1
else
  command_with_output
fi

It's pretty ugly and has repetition, so I'd like a more consice way to do it.

I tried the following, but it doesn't work.

[ -e "$1" ] && outfile=$1 || outfile='&1'
command_with_output >$outfile

Edit: This doesn't change the relevancy of the answers, but I realized after I made the question that touch "$1" && outfile=$1 is really what I need instead of [ -e "$1" ] && outfile=$1 since the file may not already exist, and I want to make sure I can write to it or create it, not just that it exists. I'm not changing the question because it would put the answers out of sync.

Answer 1

Another way is to use $1 as the filename if it was passed, and /dev/stdout otherwise (which is a symbolic link to /proc/self/fd/1 under Linux, and a device node with the same meaning on many other unix variants). E.g. put this at the top of the script:

if [ -e "$1" ]; then
    filename=$1
else
    filename=/dev/stdout
fi

And then redirect the output of every command to $filename

Answer 2

If it doesn't take any but file parameters, this will work too:

command_with_output > ${1:-/dev/stdout}

edit: Or better yet, since you probably care about errors too:

command_with_output &> ${1:-/dev/stdout}

Answer 3

exec can be used to redirect the current script's stdout to another file.

[ -e "$1" ] && exec > $1
command_with_output