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  • I understand pushd and popd to work as you describe and that they're independent of command substitution - no confusion there! However, you answered by own problem by revealing $OLDPWD. I can do popd; rmdir $OLDPWD. That;s the answer to my problem - everything else just confirms what I thought. Command substitution appears to be a way to solve it but it isn't because of the subshell and you can't do command substitution without a subshell, So thanks for revealing OLDPWD - that's exactly what I need! Commented Aug 3, 2016 at 14:47
  • @starfry, rmdir $(popd) causes popd to run in a sub-shell which means it won't change the current directory, but even if it did not run in a subshell, the output of popd will not be the temporary directory, it will be a space separated list of directories not including that temp dir. Which is where I'm saying you're confused. Commented Aug 3, 2016 at 14:59
  • but I thought I had said that in my question: "popd doesn't return the popped directory (it returns the new, now current, directory) and also because it's performed in a subshell." Admittedly it returns a list, but the first in the list is the one I was referring to. Commented Aug 3, 2016 at 15:57
  • @starfry, OK sorry. Let's say I was the one confused by the title of the question and didn't read the rest carefully enough. Commented Aug 3, 2016 at 17:30
  • well your answer was still good and pointed me in the right direction. I never knew about that shell variable OLDPWD. I must remember to one day read man bash from end-to-end ! Commented Aug 3, 2016 at 19:50