Re: sourcing script file from inside a function doesn't work anymore
On Sat, May 10, 2014 at 12:22:40PM +0200, thioroup8 wrote:
Now, I source it from inside a function: I obtain following error message:
bash: declare: tmp: not found
I think this problem raises in the particular case of involving array
variables definitions inside sourced bash script file...
cat EOF ${file_to_be_sourced}
declare -a tmp='([0]=1 [1]=2)'
EOF
The problem is that you are using declare within the file-to-be-sourced.
When you source the file inside a function, Bash runs the declare command
within the context of the function, where declare has the same meaning as
local. Thus, it makes the variable local to the function.
If you only put tmp=(...) in the file, it works as you expect. Which
is fine as long as you're only using normal arrays, and not associative
arrays.
If you want to declare associative arrays in this way, you'll need to
add the -g option to declare, which is available in bash 4.2 and later.
This makes the variable global instead of local.
There is unfortunately no middle ground between local and variable if
you must use declare (i.e. there is no way to mimic tmp=(foo bar)
to use default scoping with an associative array because you can't skip
the declare -A step on those), nor is there a way to mimic declare -g -A
in bash 4.0 or 4.1.
The only other option is to issue the declare -A outside of the function
(either in global scope, or in the outermost function where the variable
needs to be available), and then use tmp=(...) (without a declare) in
the file-to-be-sourced.
Re: sourcing script file from inside a function doesn't work anymore
2014-05-12 14:11 GMT+02:00 Greg Wooledge wool...@eeg.ccf.org:
On Sat, May 10, 2014 at 12:22:40PM +0200, thioroup8 wrote:
Now, I source it from inside a function: I obtain following error
message:
bash: declare: tmp: not found
I think this problem raises in the particular case of involving array
variables definitions inside sourced bash script file...
cat EOF ${file_to_be_sourced}
declare -a tmp='([0]=1 [1]=2)'
EOF
The problem is that you are using declare within the file-to-be-sourced.
When you source the file inside a function, Bash runs the declare command
within the context of the function, where declare has the same meaning as
local. Thus, it makes the variable local to the function.
There really is a bug here, a weird one. Using a simplified script, we
see that this works as expected:
$ source (printf declare -a array='(x)'; declare -p array\n)
declare -a array='([0]=x)'
Wrapping it in a function should yield the same output, but doesn't:
$ f() { source (printf declare -a array='(x)'; declare -p array\n); }; f
bash: declare: array: not found
And while declare -p array fails to find the array, declare -p
(without a name) does list it:
$ f() { source (printf declare -a array='(x)'; declare -p array; declare
-p | grep array= \n); }; f
bash: declare: array: not found
declare -a array='([0]=x)'
Oddly, the quotes seem to matter; changing array='(x)' to array=(x)
makes it work...
$ f() { source (printf declare -a array=(x); declare -p array\n); }; f
declare -a array='([0]=x)'
--
Geir Hauge
Re: compgen -C not working as I expected
On 5/10/14, 12:45 AM, Curious Costanza wrote: Hey Chet, Thank you for your response. I don't suppose that you have a small example handy which demonstrates a useful application of compgen with the -C or -F options, do you? I guess I still have the same question as the commenter who responded to my StackExchange post; why are -C and -F defined against compgen at all? I don't have an example, since its intent is to be able to use binaries or scripts written in a different language to produce completions. -- ``The lyf so short, the craft so long to lerne.'' - Chaucer ``Ars longa, vita brevis'' - Hippocrates Chet Ramey, ITS, CWRUc...@case.eduhttp://cnswww.cns.cwru.edu/~chet/
Re: sourcing script file from inside a function doesn't work anymore
On Mon, May 12, 2014 at 03:33:34PM +0200, Geir Hauge wrote:
Oddly, the quotes seem to matter; changing array='(x)' to array=(x)
makes it work...
$ f() { source (printf declare -a array=(x); declare -p array\n); }; f
declare -a array='([0]=x)'
OK, this also happens on my system. In fact this is the first time I've
ever seen array='(...)' instead of array=(...) and I had no idea the
quoted version works outside of a function. It seems like an incredibly
strange syntax; without the explicit declare -a it would be initializing
a string variable.
But this doesn't change the fact that the original poster is making a
local variable inside the function, which won't be visible outside it.
Re: sourcing script file from inside a function doesn't work anymore
On 5/12/14, 9:33 AM, Geir Hauge wrote:
2014-05-12 14:11 GMT+02:00 Greg Wooledge wool...@eeg.ccf.org:
On Sat, May 10, 2014 at 12:22:40PM +0200, thioroup8 wrote:
Now, I source it from inside a function: I obtain following error
message:
bash: declare: tmp: not found
I think this problem raises in the particular case of involving array
variables definitions inside sourced bash script file...
cat EOF ${file_to_be_sourced}
declare -a tmp='([0]=1 [1]=2)'
EOF
The problem is that you are using declare within the file-to-be-sourced.
When you source the file inside a function, Bash runs the declare command
within the context of the function, where declare has the same meaning as
local. Thus, it makes the variable local to the function.
There really is a bug here, a weird one. Using a simplified script, we
see that this works as expected:
Yep. Thanks for the report. The issue is that the `placeholder' attribute
doesn't get unset when the argument is quoted -- it takes a different code
path than when it's unquoted. I've attached a patch for people to
experiment with.
Chet
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRUc...@case.eduhttp://cnswww.cns.cwru.edu/~chet/
*** ../bash-4.3-patched/arrayfunc.c 2014-03-28 10:54:21.0 -0400
--- arrayfunc.c 2014-05-12 11:19:00.0 -0400
***
*** 180,183
--- 180,184
FREE (newval);
+ VUNSETATTR (entry, att_invisible); /* no longer invisible */
return (entry);
}
Re: sourcing script file from inside a function doesn't work anymore
Hi all,
Chet, I compiled a 4.3.11 from scratch.
Then I applied your fix and compiled again.
I executed my test program I attached in my bug report and it works like
a charm and like bash 4.2.
Now I'm going to try with several scripts I wrote which were affected by
this bug...
Thank you all for your time
Best regards
Jean-Philippe
Le 12/05/2014 17:31, Chet Ramey a écrit :
On 5/12/14, 9:33 AM, Geir Hauge wrote:
2014-05-12 14:11 GMT+02:00 Greg Wooledge wool...@eeg.ccf.org:
On Sat, May 10, 2014 at 12:22:40PM +0200, thioroup8 wrote:
Now, I source it from inside a function: I obtain following error
message:
bash: declare: tmp: not found
I think this problem raises in the particular case of involving array
variables definitions inside sourced bash script file...
cat EOF ${file_to_be_sourced}
declare -a tmp='([0]=1 [1]=2)'
EOF
The problem is that you are using declare within the file-to-be-sourced.
When you source the file inside a function, Bash runs the declare command
within the context of the function, where declare has the same meaning as
local. Thus, it makes the variable local to the function.
There really is a bug here, a weird one. Using a simplified script, we
see that this works as expected:
Yep. Thanks for the report. The issue is that the `placeholder' attribute
doesn't get unset when the argument is quoted -- it takes a different code
path than when it's unquoted. I've attached a patch for people to
experiment with.
Chet
