Re: No word splitting for assignment-like expressions in compound assignment
On 7/28/20 4:14 PM, Ilkka Virta wrote:
> On 28.7. 17:22, Chet Ramey wrote:
>> On 7/23/20 8:11 PM, Alexey Izbyshev wrote:
>>> $ Z='a b'
>>> $ A=(X=$Z)
>>> $ declare -p A
>>> declare -a A=([0]="X=a b")
>
>> It's an assignment statement in a context where assignment statements are
>> accepted (which is what makes it different from `echo X=$Z', for instance),
>> but the lack of a subscript on the lhs makes it a special case. I'll take a
>> look at the semantics here.
>
> This is also a bit curious:
>
> $ b=( [123]={a,b,c}x )
> $ declare -p b
> declare -a b=([0]="[123]=ax" [1]="[123]=bx" [2]="[123]=cx")
>
> It does seem to have a subscript on the LHS, but it didn't work as one.
> To be in line with a plain scalar assignment, the braces should probably
> not be expanded here.
Thanks, I'll look at it.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRUc...@case.eduhttp://tiswww.cwru.edu/~chet/
Re: No word splitting for assignment-like expressions in compound assignment
On 28.7. 17:22, Chet Ramey wrote:
On 7/23/20 8:11 PM, Alexey Izbyshev wrote:
$ Z='a b'
$ A=(X=$Z)
$ declare -p A
declare -a A=([0]="X=a b")
It's an assignment statement in a context where assignment statements are
accepted (which is what makes it different from `echo X=$Z', for instance),
but the lack of a subscript on the lhs makes it a special case. I'll take a
look at the semantics here.
This is also a bit curious:
$ b=( [123]={a,b,c}x )
$ declare -p b
declare -a b=([0]="[123]=ax" [1]="[123]=bx" [2]="[123]=cx")
It does seem to have a subscript on the LHS, but it didn't work as one.
To be in line with a plain scalar assignment, the braces should probably
not be expanded here.
--
Ilkka Virta / itvi...@iki.fi
Re: No word splitting for assignment-like expressions in compound assignment
On 7/23/20 8:11 PM, Alexey Izbyshev wrote: > Hello! > > I have a question about the following behavior: > > $ Z='a b' > $ A=(X=$Z) > $ declare -p A > declare -a A=([0]="X=a b") > $ A=(X$Z) > $ declare -p A > declare -a A=([0]="Xa" [1]="b") > > I find it surprising that no word splitting is performed in the first > compound assignment. I realize that skipping word splitting may be > desirable if a subscript is given (e.g. "A=([0]=$Z)") to make it consistent > with normal variable assignment[1], but in this case it looks like a bug. It's an assignment statement in a context where assignment statements are accepted (which is what makes it different from `echo X=$Z', for instance), but the lack of a subscript on the lhs makes it a special case. I'll take a look at the semantics here. -- ``The lyf so short, the craft so long to lerne.'' - Chaucer ``Ars longa, vita brevis'' - Hippocrates Chet Ramey, UTech, CWRUc...@case.eduhttp://tiswww.cwru.edu/~chet/
Re: No word splitting for assignment-like expressions in compound assignment
> On Jul 27, 2020, at 4:02 AM, Andreas Schwab wrote: > > On Jul 27 2020, Lawrence Velázquez wrote: > >> If word splitting were not performed in compound assignments, this... >> >>foo=(a b c) >> >> ...would not work. > > This is not true. Field splitting is only relevant for words generated > by other expansions, not for literal tokens. My fault, I got it mixed up with tokenization. Carry on! vq
Re: No word splitting for assignment-like expressions in compound assignment
On Mon, Jul 27, 2020 at 01:31:32AM -0400, Dale R. Worley wrote: > So it seems like the word splitting in "A=(X$Z)" is incorrect. If the documentation doesn't support word splitting in that case, then it's the documentation that will need to change, not the shell. Word splitting in that context is *frequently* used in scripts. Never mind the fact that it's dangerous and wrong -- people use it, a lot. A change to the shell that would stop it from occurring would break countless scripts.
Re: No word splitting for assignment-like expressions in compound assignment
27 Temmuz 2020 Pazartesi tarihinde Alexey Izbyshev yazdı: > On 2020-07-27 10:06, Lawrence Velázquez wrote: > >> On Jul 27, 2020, at 1:31 AM, Dale R. Worley wrote: >>> Interesting. The documentation for 4.2.53(1) says this about parameter >>> assignments generally, with no special rules for compound assignments: >>> >>> All >>> values undergo tilde expansion, parameter and variable expansion, >>> com- >>> mand substitution, arithmetic expansion, and quote removal (see >>> EXPAN- >>> SION below). ... Word splitting is not >>> performed, with the exception of "$@" as explained below under >>> Special >>> Parameters. Pathname expansion is not performed. >>> >>> So it seems like the word splitting in "A=(X$Z)" is incorrect. So is >>> pathname expansion in that context. >>> >> >> >> If word splitting were not performed in compound assignments, this... >> >> foo=(a b c) >> >> ...would not work. If pathname expansion were not performed in compound >> assignments, this... >> >> foo=(*) >> >> ...would not work. Arrays would become significantly less usable if >> word splitting and pathname expansion were not allowed in compound >> assignments. >> >> To be clear, I don't consider word splitting and expansions in compound > assignments to be a problem: this is well-known and long-standing behavior, > even though it doesn't seem to be explicitly documented. In particular, I > expect word splitting to happen in "A=(X$Z)" case. But I expect it to > happen in "A=(X=$Z)" too, and the lack of it seems unintentional to me. I agree, anything that forms a valid assignment statement in isolation is exempt from word splitting and that indeed seems like a bug or very poor implementation choice. $ Z='a b' $ A=(X=$Z X[123]=$Z X[qwerty]=$Z X+=$Z) $ declare -p A declare -a A=([0]="X=a b" [1]="X[123]=a b" [2]="X[qwerty]=a b" [3]="X+=a b") > > Alexey > > -- Oğuz
Re: No word splitting for assignment-like expressions in compound assignment
On Jul 27 2020, Lawrence Velázquez wrote: > If word splitting were not performed in compound assignments, this... > > foo=(a b c) > > ...would not work. This is not true. Field splitting is only relevant for words generated by other expansions, not for literal tokens. Andreas. -- Andreas Schwab, sch...@linux-m68k.org GPG Key fingerprint = 7578 EB47 D4E5 4D69 2510 2552 DF73 E780 A9DA AEC1 "And now for something completely different."
Re: No word splitting for assignment-like expressions in compound assignment
On 2020-07-27 10:06, Lawrence Velázquez wrote: On Jul 27, 2020, at 1:31 AM, Dale R. Worley wrote: Interesting. The documentation for 4.2.53(1) says this about parameter assignments generally, with no special rules for compound assignments: All values undergo tilde expansion, parameter and variable expansion, com- mand substitution, arithmetic expansion, and quote removal (see EXPAN- SION below). ... Word splitting is not performed, with the exception of "$@" as explained below under Special Parameters. Pathname expansion is not performed. So it seems like the word splitting in "A=(X$Z)" is incorrect. So is pathname expansion in that context. If word splitting were not performed in compound assignments, this... foo=(a b c) ...would not work. If pathname expansion were not performed in compound assignments, this... foo=(*) ...would not work. Arrays would become significantly less usable if word splitting and pathname expansion were not allowed in compound assignments. To be clear, I don't consider word splitting and expansions in compound assignments to be a problem: this is well-known and long-standing behavior, even though it doesn't seem to be explicitly documented. In particular, I expect word splitting to happen in "A=(X$Z)" case. But I expect it to happen in "A=(X=$Z)" too, and the lack of it seems unintentional to me. Alexey
Re: No word splitting for assignment-like expressions in compound assignment
> On Jul 27, 2020, at 1:31 AM, Dale R. Worley wrote:
>
> Alexey Izbyshev writes:
>> I have a question about the following behavior:
>>
>> $ Z='a b'
>> $ A=(X=$Z)
>> $ declare -p A
>> declare -a A=([0]="X=a b")
>> $ A=(X$Z)
>> $ declare -p A
>> declare -a A=([0]="Xa" [1]="b")
>>
>> I find it surprising that no word splitting is performed in the first
>> compound assignment.
>
>> * Brace expansion is performed for "A=(X=a{x,y}b)" by all bash versions
>> mentioned above (which is inconsistent with normal variable assignment).
>> * Globbing for "A=(X=a?b)" is performed by bash 3.1.17, but not by other
>> versions.
>
> Interesting. The documentation for 4.2.53(1) says this about parameter
> assignments generally, with no special rules for compound assignments:
>
> All
> values undergo tilde expansion, parameter and variable expansion, com-
> mand substitution, arithmetic expansion, and quote removal (see EXPAN-
> SION below). ... Word splitting is not
> performed, with the exception of "$@" as explained below under Special
> Parameters. Pathname expansion is not performed.
>
> So it seems like the word splitting in "A=(X$Z)" is incorrect. So is
> pathname expansion in that context.
If word splitting were not performed in compound assignments, this...
foo=(a b c)
...would not work. If pathname expansion were not performed in compound
assignments, this...
foo=(*)
...would not work. Arrays would become significantly less usable if
word splitting and pathname expansion were not allowed in compound
assignments.
vq
Re: No word splitting for assignment-like expressions in compound assignment
Alexey Izbyshev writes:
> I have a question about the following behavior:
>
> $ Z='a b'
> $ A=(X=$Z)
> $ declare -p A
> declare -a A=([0]="X=a b")
> $ A=(X$Z)
> $ declare -p A
> declare -a A=([0]="Xa" [1]="b")
>
> I find it surprising that no word splitting is performed in the first
> compound assignment.
> * Brace expansion is performed for "A=(X=a{x,y}b)" by all bash versions
> mentioned above (which is inconsistent with normal variable assignment).
> * Globbing for "A=(X=a?b)" is performed by bash 3.1.17, but not by other
> versions.
Interesting. The documentation for 4.2.53(1) says this about parameter
assignments generally, with no special rules for compound assignments:
All
values undergo tilde expansion, parameter and variable expansion, com-
mand substitution, arithmetic expansion, and quote removal (see EXPAN-
SION below). ... Word splitting is not
performed, with the exception of "$@" as explained below under Special
Parameters. Pathname expansion is not performed.
So it seems like the word splitting in "A=(X$Z)" is incorrect. So is
pathname expansion in that context. Oddly, brace expansion is not
mentioned.
Dale
No word splitting for assignment-like expressions in compound assignment
Hello!
I have a question about the following behavior:
$ Z='a b'
$ A=(X=$Z)
$ declare -p A
declare -a A=([0]="X=a b")
$ A=(X$Z)
$ declare -p A
declare -a A=([0]="Xa" [1]="b")
I find it surprising that no word splitting is performed in the first
compound assignment. I realize that skipping word splitting may be
desirable if a subscript is given (e.g. "A=([0]=$Z)") to make it
consistent with normal variable assignment[1], but in this case it looks
like a bug.
I've reproduced the described behavior on bash 4.4.20 (Ubuntu 18.04),
3.1.17 (CentOS 5) and self-built 5.1-alpha release.
Some other discoveries:
* Brace expansion is performed for "A=(X=a{x,y}b)" by all bash versions
mentioned above (which is inconsistent with normal variable assignment).
* Globbing for "A=(X=a?b)" is performed by bash 3.1.17, but not by other
versions.
[1] https://lists.gnu.org/archive/html/bug-bash/2012-08/msg00055.html
Alexey
