On Thu, Oct 26, 2006 at 12:14:16PM -0400, A. Costa wrote:
On Mon, 23 Oct 2006 10:05:39 +
Gerrit Pape [EMAIL PROTECTED] wrote:
...Here bash --posix behaves differently in the first case.
Yet another view, without the '-x', plus a few more shells:
% for f in dash posh ksh
On Mon, 23 Oct 2006 10:05:39 +
Gerrit Pape [EMAIL PROTECTED] wrote:
...Here bash --posix behaves differently in the first case.
Yet another view, without the '-x', plus a few more shells:
% for f in dash posh ksh pdksh bash bash --posix; do echo -n $f:
; $f -c 'export x=$@;
On Sat, Oct 21, 2006 at 02:37:25PM -0400, A. Costa wrote:
I'm about to close this bug, thanks, Gerrit.
Thanks for the detailed explanation, and the good advice how not to get
the error using $*. And apologies for my misunderstanding... but in
hopes of extracting some public good from it, a
On Mon, 23 Oct 2006 08:09:48 +
Gerrit Pape [EMAIL PROTECTED] wrote:
If 'dash' and 'posh' behave correctly, and 'bash' does it
differently, would that indicate a bug in how 'bash' parses $@?
The 'bash'
Yes.
Then would you prefer this bug be reassigned to 'bash'?
I'd expect x=$@ to
On Mon, Oct 23, 2006 at 05:12:11AM -0400, A. Costa wrote:
On Mon, 23 Oct 2006 08:09:48 +
Gerrit Pape [EMAIL PROTECTED] wrote:
If 'dash' and 'posh' behave correctly, and 'bash' does it
differently, would that indicate a bug in how 'bash' parses $@?
The 'bash'
Yes.
Then would
On Wed, Aug 02, 2006 at 01:20:44AM -0400, A. Costa wrote:
In 'dash' try this:
% foo() { export x=$@ ; }
% foo -f --c
export: 4: --c: bad variable name
% echo $?
2
It seems like it should be standard code.
Remove 'export' from 'foo()' and
On Sat, 21 Oct 2006 14:36:04 +
Gerrit Pape [EMAIL PROTECTED] wrote:
Hi A., I don't think dash and posh are at fault here...
...This means export x=$@ with $*=-f --c will expand to export x=-f
--c, and dash's error message is quite right. bash adds quotes in
this case, but you actually
Package: dash
Version: 0.5.3-3
Severity: normal
In 'dash' try this:
% foo() { export x=$@ ; }
% foo -f --c
export: 4: --c: bad variable name
% echo $?
2
It seems like it should be standard code.
Remove 'export' from 'foo()' and there's no error:
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