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Add a missing "$" on a variable usage #908

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Add a missing "$" on a variable usage #908

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2 changes: 1 addition & 1 deletion highlighters/main/main-highlighter.zsh
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Original file line number Diff line number Diff line change
Expand Up @@ -1537,7 +1537,7 @@ _zsh_highlight_main_highlighter_highlight_double_quote()
(( i += 2 ))
_zsh_highlight_main_highlighter_highlight_list $(( start_pos + i - 1 )) S $has_end $arg[i,-1]
ret=$?
(( i += REPLY ))
(( i += $REPLY ))
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Thanks for the patch. Unfortunately, it's not a complete fix.

Arithmetic expressions are parsed for substitutions and then, in a second pass, evaluate identifiers into their values. Thus, the incumbent code is correct, and even idiomatic (consider foo=2+2; echo $((foo*2)) $(($foo*2))), and conversely, the patch would seem to merely paper over the issue.

I'm curious why it works, but we can discuss that back on the issue's ticket.

last_break=$(( start_pos + i ))
reply=(
$saved_reply
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