On Tue, Oct 20, 2009 at 10:17 PM, Dan McGee wrote:

On Tue, Oct 20, 2009 at 9:45 PM, Allan McRae wrote:

...

Hi all,

This problem has been doing my head in...    First a minimal example that reflects how makepkg does things:

--one.sh-- #!/bin/bash

echo "pass 1:" for arg in "$@"; do  echo $arg done echo

ARGLIST="$@"

./two.sh $ARGLIST --end one.sh--

--two.sh-- #!/bin/bash

echo "pass 2:" for arg in "$@"; do  echo $arg done --end two.sh--

then run: ./one.sh -f -h "foo bar" pass 1: -f -h foo bar

pass 2: -f -h foo bar

Note how in pass two, foo and bar are no longer in the one line.   Of course, passing ./two.sh "$@" works, but the argument parsing in makepkg clears that, hence the need to save it to ARGLIST.

Any ideas?

Of course! I think I got it.

dmcgee@kilkenny /tmp $ ./one.sh -f -h "foo bar" pass 1: -f -h foo bar

pass 2: -f -h foo bar

dmcgee@kilkenny /tmp $ ./one-new.sh -f -h "foo bar" pass 1: -f -h foo bar

pass 2: -f -h foo bar

$ cat one-new.sh #!/bin/bash echo "pass 1:" for arg in "$@"; do  echo $arg done echo

ARGLIST=("$@")

./two.sh "${ARGLIST[@]}"

Do I win a prize or anything? :P

It's worth noting that the following works two: ./two.sh "$@" The reason being that "$@" is special in bash. It actually expands to the command line quoted as you passed it, with "foo bar" being one argument. The issue is with the assignment to a single bash variable

Aaron Griffin wrote:

On Tue, Oct 20, 2009 at 10:17 PM, Dan McGee wrote:

...

On Tue, Oct 20, 2009 at 9:45 PM, Allan McRae wrote:

...

Hi all,

This problem has been doing my head in... First a minimal example that reflects how makepkg does things:

--one.sh-- #!/bin/bash

echo "pass 1:" for arg in "$@"; do echo $arg done echo

ARGLIST="$@"

./two.sh $ARGLIST --end one.sh--

--two.sh-- #!/bin/bash

echo "pass 2:" for arg in "$@"; do echo $arg done --end two.sh--

then run: ./one.sh -f -h "foo bar" pass 1: -f -h foo bar

pass 2: -f -h foo bar

Note how in pass two, foo and bar are no longer in the one line. Of course, passing ./two.sh "$@" works, but the argument parsing in makepkg clears that, hence the need to save it to ARGLIST.

Any ideas?

Of course! I think I got it.

dmcgee@kilkenny /tmp $ ./one.sh -f -h "foo bar" pass 1: -f -h foo bar

pass 2: -f -h foo bar

dmcgee@kilkenny /tmp $ ./one-new.sh -f -h "foo bar" pass 1: -f -h foo bar

pass 2: -f -h foo bar

$ cat one-new.sh #!/bin/bash echo "pass 1:" for arg in "$@"; do echo $arg done echo

ARGLIST=("$@")

./two.sh "${ARGLIST[@]}"

Do I win a prize or anything? :P

It's worth noting that the following works two:

./two.sh "$@"

The reason being that "$@" is special in bash. It actually expands to the command line quoted as you passed it, with "foo bar" being one argument. The issue is with the assignment to a single bash variable

The reason that can not be used in makepkg is the option parsing uses "shift" and thus clears the value of $@ as it goes. Allan

On Tue, Oct 20, 2009 at 10:34 PM, Dan McGee wrote:

On Tue, Oct 20, 2009 at 10:30 PM, Allan McRae wrote:

...

Aaron Griffin wrote:

...

On Tue, Oct 20, 2009 at 10:17 PM, Dan McGee wrote:

...

On Tue, Oct 20, 2009 at 9:45 PM, Allan McRae wrote:

...

Note how in pass two, foo and bar are no longer in the one line.   Of course, passing ./two.sh "$@" works, but the argument parsing in makepkg clears that, hence the need to save it to ARGLIST.

It's worth noting that the following works two:

./two.sh "$@"

The reason being that "$@" is special in bash. It actually expands to the command line quoted as you passed it, with "foo bar" being one argument. The issue is with the assignment to a single bash variable

The reason that can not be used in makepkg is the option parsing uses "shift" and thus clears the value of $@ as it goes.

I thought I'd craft a witty reply but you beat me with a nicely worded one. I happily trimmed the reply chain above to see the failure to read. :P

Whoops. I didn't see it mentioned. I did, however, want to point out the fact that "$@" is special. It's not just a variable in quotes. It's crazy like that