odd `shift' behaviour with bash-1.12

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2 total messages Started by Salvador Pinto Abreu Thu, 27 Aug 1992 14:36

#916

Author: Salvador Pinto Abreu
Date: Thu, 27 Aug 1992 14:36

33 lines
1398 bytes

The scope of the `shift' command seems to be the innermost loop. For
example, the following code:

        MAX_PARALLEL=...
        MAX_PARALLEL_COUNT=$(echo -n $MAX_PARALLEL | wc -c)

        function bug {
          local PARALLEL

          while [ -n "$1" ]; do
            PARALLEL=$MAX_PARALLEL
            echo 1>&2 bug: entering parallel loop.
            while [ -n "$1" -a -n "$PARALLEL" ]; do
              ( echo bug: output of $1 & )
              PARALLEL=${PARALLEL#.}
              shift
            done | cat
            [ -z "$LOOP" ] && shift $MAX_PARALLEL_COUNT
          done
        }

        bug 1 2 3 4 5 6 7 8	# runs fine
        LOOP=on
	bug 1 2 3 4 5 6 7 8	# loops

the second `shift' needs to be executed to shift the args to `bug',
even though they were already (supposed to have been) kicked out by
the `shift' in the inner loop.

bug or feature?

PS: tried on Ultrix 4.2, HP/UX A.B8.05 and NeXT Mach 2.2.
--

Re: odd `shift' behaviour with bash-1.12

#921

Author: Robert Elz
Date: Sat, 29 Aug 1992 17:05

15 lines
1112 bytes

In <SPA.92Aug27153635@khore-dump.fct.unl.pt>
spa@fct.unl.PT (Salvador Pinto Abreu) writes:

| The scope of the `shift' command seems to be the innermost loop. For
| example, the following code:

Take away the "| cat" which takes the output from your inner loop,
and try it then (though your output will no longer be buffered by cat).

Its SOP for shells that when a control statement has I/O redirected
the control statement is run in a sub-shell, and so can't affect anything
in the environment of the real shell (including $# etc).  Just why it
has to be that way has never been clear to me, but that's the way it is.

kre

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