You can also fool sh into running the *wrong* binary if if you have
two in showdowed paths:
#! /bin/sh
test -d foo1 || mkdir foo1
test -d foo2 || mkdir foo2
test -d foo2 || mkdir foo3
echo 'echo :one' > foo1/run
echo 'echo :two' > foo2/run
echo 'echo :three' > foo2/run3
chmod a+x */run*
hash -r
echo
echo Expect one:
PATH=./foo3:./foo1:./foo2:./foo5
echo Expect two:
PATH=./foo3:./foo3:./foo1 run
echo run should be in in foo1:
hash -v
echo $PATH
echo Should give one:
run
==> runs foo2/run, not foo1/run.
This is still covered by the quick fix I sent. Looking for cases that
aren't...
Martin
--
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Martin Cracauer <[EMAIL PROTECTED]> http://www.cons.org/cracauer/
BSD User Group Hamburg, Germany http://www.bsdhh.org/
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