For some reason, I am not able to access variables that were changed in a
while loop, after the while loop finishes. For example:

i=0
echo "this is a test" | while read line
do
i = `expr $i + 1`
done
echo $i

-------------------------------
The echo statement prints 0 rather than 1. Does anyone know why?

"Doug Ritschel" <ritschel@optonline.net> writes:

> i=0
> echo "this is a test" | while read line; do
> i = `expr $i + 1`
> done
> echo $i

The while loop is run in a subshell and it has it's own local
variables so changing them won't change the value of oryginal
variable.

In fact AFAIK the question was answered in FAQ.


--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl>--<jid:mina86*jabber.org>--ooO--(_)--Ooo--

"Michal Nazarewicz" <mina86@tlen.pl> wrote in message
news:87r6z3d27r.fsf@erwin.piotrekn...
> "Doug Ritschel" <ritschel@optonline.net> writes:
>
> > i=0
> > echo "this is a test" | while read line; do
> > i = `expr $i + 1`
> > done
> > echo $i
>
> The while loop is run in a subshell and it has it's own local
> variables so changing them won't change the value of oryginal
> variable.
>
> In fact AFAIK the question was answered in FAQ.
>
>

Why, then , if I run the following, the echo prints 3, which is the number
of lines in the hosts file.

i=0
while read line; do
i = `expr $i + 1`
done < /etc/hosts
echo $i

"Doug Ritschel" <ritschel@optonline.net> wrote in message
news:8sZHg.206$Y43.108@newsfe09.lga...
>
> "Michal Nazarewicz" <mina86@tlen.pl> wrote in message
> news:87r6z3d27r.fsf@erwin.piotrekn...
> > "Doug Ritschel" <ritschel@optonline.net> writes:
> >
> > > i=0
> > > echo "this is a test" | while read line; do
> > > i = `expr $i + 1`
> > > done
> > > echo $i
> >
> > The while loop is run in a subshell and it has it's own local
> > variables so changing them won't change the value of oryginal
> > variable.
> >
> > In fact AFAIK the question was answered in FAQ.
> >
> >
>
> Why, then , if I run the following, the echo prints 3, which is the number
> of lines in the hosts file.
>
> i=0
> while read line; do
> i = `expr $i + 1`
> done < /etc/hosts
> echo $i
>
>

Found the below answer:

The disadvantage of the pipe-to-while method is that each element in the
pipeline is run in a subshell, so variables set inside the while loop
(for example) won't be set after the loop is complete[*].

find . | {
while read line
do
: whatever
word=${line%% *}
done

: variables set within the loop are still available here, e.g.:
printf "%s\n" "$word"

}

"Doug Ritschel" <ritschel@optonline.net> writes:

> "Michal Nazarewicz" <mina86@tlen.pl> wrote in message
> news:87r6z3d27r.fsf@erwin.piotrekn...
>> "Doug Ritschel" <ritschel@optonline.net> writes:
>>
>> > i=0
>> > echo "this is a test" | while read line; do
>> > i = `expr $i + 1`
>> > done
>> > echo $i
>>
>> The while loop is run in a subshell and it has it's own local
>> variables so changing them won't change the value of oryginal
>> variable.
>
> Why, then , if I run the following, the echo prints 3, which is the
> number of lines in the hosts file.
>
> i=0
> while read line; do i=`expr $i + 1`; done < /etc/hosts
> echo $i

Because there is no need for subshell. In the first while loop you've
used a pipe so shell had to run another process. In the second while
loop you didn't use a pipe but only redirect stdout to be read from
/etc/hosts.

--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl>--<jid:mina86*jabber.org>--ooO--(_)--Ooo--

On 2006-08-26, Doug Ritschel wrote:
>
> "Michal Nazarewicz" <mina86@tlen.pl> wrote in message
> news:87r6z3d27r.fsf@erwin.piotrekn...
>> "Doug Ritschel" <ritschel@optonline.net> writes:
>>
>> > i=0
>> > echo "this is a test" | while read line; do
>> > i = `expr $i + 1`
>> > done
>> > echo $i
>>
>> The while loop is run in a subshell and it has it's own local
>> variables so changing them won't change the value of oryginal
>> variable.
>>
>> In fact AFAIK the question was answered in FAQ.
>>
>>
>
> Why, then , if I run the following, the echo prints 3, which is the number
> of lines in the hosts file.
>
> i=0
> while read line; do
> i = `expr $i + 1`
> done < /etc/hosts
> echo $i

First, you did not get any output from that script unless you have
an executable named 'i'. You are asking the shell to execute a
command named 'i', with arguments of '=' and '1'; you are not
incrementing a variable. You meant: i=`expr $i + 1` (no spaces
around the equals sign).

Second, you were not running it (with the error fixed) in a Bourne
shell. If you were, you would still have seen '0'. In a Bourne
shell, the redirecion alone is enough to force a subshell.


--
Chris F.A. Johnson, author <http://cfaj.freeshell.org>
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
===== My code in this post, if any, assumes the POSIX locale
===== and is released under the GNU General Public Licence

"Chris F.A. Johnson" <cfajohnson@gmail.com> writes:

> On 2006-08-26, Doug Ritschel wrote:
>> i=0
>> while read line; do
>> i = `expr $i + 1`
>> done < /etc/hosts
>> echo $i
>
> Second, you were not running it (with the error fixed) in a Bourne
> shell. If you were, you would still have seen '0'. In a Bourne
> shell, the redirecion alone is enough to force a subshell.

Are you sure?

$ cat test.sh
i=0
while read line; do i=$(($i + 1)); done </etc/hosts
echo $i
$ ash test.sh
24
$ bash test.sh
24
$ zsh test.sh
24


Any pointers to standards which support your claim?

--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl>--<jid:mina86*jabber.org>--ooO--(_)--Ooo--

2006-08-27, 13:57(+02), Michal Nazarewicz:
> "Chris F.A. Johnson" <cfajohnson@gmail.com> writes:
>
>> On 2006-08-26, Doug Ritschel wrote:
>>> i=0
>>> while read line; do
>>> i = `expr $i + 1`
>>> done < /etc/hosts
>>> echo $i
>>
>> Second, you were not running it (with the error fixed) in a Bourne
>> shell. If you were, you would still have seen '0'. In a Bourne
>> shell, the redirecion alone is enough to force a subshell.
>
> Are you sure?
>
> $ cat test.sh
> i=0
> while read line; do i=$(($i + 1)); done </etc/hosts
> echo $i
> $ ash test.sh
> 24
> $ bash test.sh
> 24
> $ zsh test.sh
> 24
>
>
> Any pointers to standards which support your claim?
[...]

Neither ash, bash or zsh are the Bourne shell. Your ash is not
the original ash as it supports $((...)). You can find a Bourne
shell as the /bin/sh of Solaris or some very old commercial
Unices.

The Bourne shell doesn't have $((...)), and there's no standard
specifying the Bourne shell.

In the Bourne shell, redirection meant fork. Newer shells use
some trick to work around that limitation (they save the
redirected fds (for restauration afterwards) by dupplicating
them on other fds before doing the redirections, which you could
do manually with the Bourne shell this way:
exec 3<&0 < /etc/hosts
while read line
do i=`expr "$i" + 1 3<&-`
done
exec <&3 3<&-
)


--
Stéphane