Re: $var not expanded in ${x?$var}

[Stephane Chazelas <stephane@xxxxxxxxxxxx>]

2023-01-16 17:15:35 +0000, Peter Stephenson:
> > On 13/01/2023 08:02 Stephane Chazelas <stephane@xxxxxxxxxxxx> wrote:
> >  
> > $ zsh -c 'echo ${1?$USERNAME}'
> > zsh:1: 1: $USERNAME
> > 
> > No quote removal either:
> > 
> > $ zsh -c 'echo ${1?"x"}'
> > zsh:1: 1: "x"
> > 
> > Doc says:
> > 
> > > In any of the above expressions that test a variable and substitute an
> > > alternate WORD, note that you can use standard shell quoting in the WORD
> > > value to selectively override the splitting done by the SH_WORD_SPLIT
> > > option and the = flag, but not splitting by the s:STRING: flag.
> 
> In fact the shell does not "substitute an alternate WORD" here, it
> just prints it out, but the difference is easy to miss and expanding it
> seems the right thing to do from other points of view, so I've noted it
> in the doc.
[...]

Thanks.

$ echo ${1?$'a\nb'}
zsh: 1: a\nb
$ print -Prv err '%F{red}BAD%f'
$ a=${1?$err}
zsh: 1: ^[[31mBAD^[[39m

That transformation of newline into \n and other nicezputs'ness
seems undesirable to me though.

It doesn't seem unreasonable for someone to want to provide a
multiline error message or one with terminal escape sequences.

See for instance
https://unix.stackexchange.com/questions/769679/bash-parameter-substitution-with-multi-line-error-message

AFAICT, except for bash doing the IFS-split + join, all other
shells send the expansion verbatim to stderr.

-- 
Stephane