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



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 


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