[bug] Math evaluation done twice in subscript in specific conditions

[bug] Math evaluation done twice in subscript in specific conditions

[Sebastian Gniazdowski]

Hello
This is interesting:

idx=0
a=()
: ${a[++idx]::=$idx}
print -rl $a
-> 1
: ${a[++idx]::=$idx}
print -rl $a
-> 1
-> 3
idx=1
: ${a[++idx]::=$idx}
print -rl $a
-> 1
-> 2
-> 3

So apparently the assignment to idx does change something for the
following ::= "call" and makes it work correctly. The second and
following ::= calls cause $idx to increment twice.

-- 
Sebastian Gniazdowski
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Re: [bug] Math evaluation done twice in subscript in specific conditions

[Stephane Chazelas]

2019-07-24 01:50:23 +0200, Sebastian Gniazdowski:
[...]
> idx=0
> a=()
> : ${a[++idx]::=$idx}
> print -rl $a
> -> 1
> : ${a[++idx]::=$idx}
> print -rl $a
> -> 1
> -> 3
> idx=1
> : ${a[++idx]::=$idx}
> print -rl $a
> -> 1
> -> 2
> -> 3
> 
> So apparently the assignment to idx does change something for the
> following ::= "call" and makes it work correctly. The second and
> following ::= calls cause $idx to increment twice.
[...]

It is increased by 2 on the first one as well:

$ i=0 a=()
$ : ${a[++i]::=$i}; echo $i $a
2 1
$ : ${a[++i]::=$i}; echo $i $a
4 1 3

With:

$ i='psvar[0`uname>&2`]'
$ : ${a[i]}
Linux
$ : ${a[i]::=}
Linux
Linux
zsh: a: assignment to invalid subscript range

So the extra evaluation of the indice is done *before* the
assignment it would seem.

$ echo ${a[i]:=x}
Linux
Linux
zsh: a: assignment to invalid subscript range

It makes more sense in that latter case where I imaging a[i] is
dereferenced first to check if it's empty or not, and then the
indice is computed again upon assignment. 

-- 
Stephane

Re: [bug] Math evaluation done twice in subscript in specific conditions

[Sebastian Gniazdowski]

On Wed, 24 Jul 2019 at 07:12, Stephane Chazelas
<stephane.chazelas@gmail.com> wrote:
>
> 2019-07-24 01:50:23 +0200, Sebastian Gniazdowski:
> [...]
> > idx=0
> > a=()
> > : ${a[++idx]::=$idx}
> > print -rl $a
> > -> 1
> > : ${a[++idx]::=$idx}
> > print -rl $a
> > -> 1
> > -> 3
> > idx=1
> > : ${a[++idx]::=$idx}
> > print -rl $a
> > -> 1
> > -> 2
> > -> 3
> >
> > So apparently the assignment to idx does change something for the
> > following ::= "call" and makes it work correctly. The second and
> > following ::= calls cause $idx to increment twice.
> [...]
>
> It is increased by 2 on the first one as well:
>
> $ i=0 a=()
> $ : ${a[++i]::=$i}; echo $i $a
> 2 1
> $ : ${a[++i]::=$i}; echo $i $a
> 4 1 3

Ah, of course.

> It makes more sense in that latter case where I imaging a[i] is
> dereferenced first to check if it's empty or not, and then the
> indice is computed again upon assignment.
When I was debugging Zsh in the past, I've very often occurred double
getter call AFAIR, or at least of something underneath. It is basic to
assume that this would call mathevali twice. I was then thinking that
it is like that by design and didn't report it.

-- 
Sebastian Gniazdowski
News: https://twitter.com/ZdharmaI
IRC: https://kiwiirc.com/client/chat.freenode.net:+6697/#zplugin
Blog: http://zdharma.org