Add a missing "$" on a variable usage #908
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@danielshahaf please review |
| (( 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.
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Fixes #907