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    $(( is not necessarily atomic. Behaviour for $((foo)|(bar)) varies with shells. Commented Jan 28, 2015 at 10:44
  • I begin the answer with "in bash:"... :) Commented Jan 28, 2015 at 11:06
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    Well, in bash, $((foo)|(bar)) does process substitution. Commented Jan 28, 2015 at 11:53