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    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. Commented Nov 25, 2011 at 2:25
  • @GordonDavisson But printargs() prints same output if no arguments or if one empty string argument. Commented Dec 17, 2019 at 13:51