The syntax isn't quite ${1:+$foo}, and hence "this" or "that". I've never seen any documentation of the syntax ${foo+"$bar"}.

I'm just curious what the 1+ does in this context. If there are actual arguments in "$@", the 1+ doesn't seem to modify the end result at all. So, if "$@" is empty, do we get instead the return value of the first positional parameter $1, which happens also to be empty? Or does the 1+ just act to clean up the return value of the empty "$@".

In reply to Re^2: what's the difference between "$@" and ${1+"$@"} by cadphile
in thread what's the difference between "$@" and ${1+"$@"} by cadphile

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