In bash, if you put:

ls /Users/*/.ssh/id_rsa 2>&1 > rsa-keys.log

…you’re redirecting stderr to the stdout’s destination while stdout is still sending output to the screen. So any permission errors encountered will go to the screen, not to rsa-keys.log.

From the bash manpage:

==================

Note that the order of redirections is significant. For example, the command

   ls > dirlist 2>&1

directs both standard output and standard error to the file dirlist, while the command

   ls 2>&1 > dirlist

directs only the standard output to file dirlist, because the standard error was duplicated from the standard output before the standard output was redirected to dirlist.

==================

Commands given to the shell are evaluated and processed in a specific order and fashion, and this is one quirk of that that many people are unaware of.

    • lousyd@lemmy.sdf.orgOP
      4·
      6 days ago

      Your way would spawn a whole extra process, but if you’re running Bash commands I suppose that doesn’t often matter.

      • INeedMana@piefed.zipEnglish
        1·
        5 days ago

        https://www.man7.org/linux/man-pages/man1/bash.1.html

        Pipelines A pipeline is a sequence of one or more commands separated by one of the control operators | or |&. The format for a pipeline is:

        [time [-p]] [ ! ] command1 [ [|⎪|&] command2 … ]

        (…) If |& is used, command1’s standard error, in addition to its standard output, is connected to command2’s standard input through the pipe; it is shorthand for 2>&1 |. This implicit redirection of the standard error to the standard output is performed after any redirections specified by command1.

        • MonkderVierte@lemmy.zip
          1·
          5 days ago

          So it is a bash thing.

          Just to be clear, because some seem to conflate Bash with Shell. If not specified, assume POSIX shell, that’s how /bin/sh is handled as well. And that has no |&.