Re: Late variable evaluation

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engeler@xxxxxxxxx wrote:
I thought the process "echo $t" is started anew on each iteration of
the while-loop, and therefore the variable $t read again. $t belongs to
the shell, not to the process?



Given your original statement of the shellcode
t=foo
( while true; do echo $t; sleep 1; done ) &
then your understanding is incorrect.

The code fragment
( while true; do echo $t; sleep 1; done )
will be launched in a subshell, and once that happens, any changes to
variables in the parent shell will not be passed on to the code fragment.

When this fragment is launched, $t has the value of 'foo', and so the subshell
inherits that value for /it's/ version of $t. If you later change the parent
shell's value for $t, the change will not (and cannot) be propogated to the
subshell running the code fragment.

- --
Lew Pitcher

Master Codewright & JOAT-in-training | GPG public key available on request
Registered Linux User #112576 (http://counter.li.org/)
Slackware - Because I know what I'm doing.
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