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Dec 17, 2019 at 13:51 comment added JohnMudd @GordonDavisson But printargs() prints same output if no arguments or if one empty string argument.
Nov 25, 2011 at 2:25 comment added Gordon Davisson echo isn't showing what you think it is. Try printargs() { printf "'%s' " "$@"; echo; }; printargs $FLAGS; printargs "$FLAGS" to see why neither of these options work.
Nov 24, 2011 at 17:49 history answered Franck Lefort CC BY-SA 3.0