Re: set -u not working as expected

[Lawrence Velázquez]

> On Aug 1, 2020, at 8:47 PM, Lawrence Velázquez <vq@larryv.me> wrote:
> 
> Presumably none of these shells implements u+=(t) as u=("${u[@]}" t).

Granted, they do disagree on ${u[@]}.

% bash -c 'set -u; unset u; u=("${u[@]}" t); typeset -p u'
declare -a u=([0]="t")
% ksh -c 'set -u; unset u; u=("${u[@]}" t); typeset -p u'
typeset -a u=(t)
% zsh -c 'set -u; unset u; u=("${u[@]}" t); typeset -p u'
zsh:1: u[@]: parameter not set

> I haven't seen the code for arithmetic expansion, but I assume it
> treats v+=1 as morally equivalent to v=${v}+1 (à la C99). Thus there
> *is* an expansion, which fails under set -u. Regardless of the
> particulars, ksh and zsh again agree:
> 
> % bash -c 'set -u; unset v; let v+=1; printf "<%s>\\n" "$v"'
> bash: v: unbound variable
> % ksh -c 'set -u; unset v; let v+=1; printf "<%s>\\n" "$v"'
> ksh: let: v: parameter not set
> ksh: v: parameter not set
> % zsh -c 'set -u; unset v; let v+=1; printf "<%s>\\n" "$v"'
> zsh:1: v: parameter not set
> zsh:1: v: parameter not set

On the other hand...

% bash -c 'set -u; typeset -i v; printf "<%s>\\n" "$v"'
bash: v: unbound variable
% ksh -c 'set -u; typeset -i v; printf "<%s>\\n" "$v"'
ksh: v: parameter not set
% zsh -c 'set -u; typeset -i v; printf "<%s>\\n" "$v"'
<0>

...and...

% bash -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
<1>
% ksh -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
<1>
% zsh -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
<1>

*shrug*

vq