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  • Hi... what is the purpose of the sed in this snippet ? Commented Jun 17, 2020 at 0:33
  • 1
    @MarcoZen it is to remove the literal $variable1 = and then the trailing ;. So you are setting the bash $variable1 to 'foo2 with bar2'. Commented Jun 22, 2020 at 20:07
  • Thanks for clarifying that ... Commented Jun 23, 2020 at 5:54