On 5/1/17 10:00 AM, Greg Wooledge wrote:
> On Mon, May 01, 2017 at 03:49:39PM +0200, Florian Mayer wrote:
>> However,
>> $ (({1..10}'+' +0))
>> Gives me "bash: ((: 1+ +0 2+ +0 3+ +0 4+ +0 5+ +0 6+ +0 7+ +0 8+ +0 9+
>> +0 10+ +0: syntax"
>> which is the same thing I'd get, when I whould've done {1..10}'+ +0'. Thus
>> in this same arithmetic expansion context bash _does_ indeed do brace
>> expansion. But it does
>> it after it deleted all whitespace inside the (( )) pair.
>>
>> Why is that?
>
> Ooh, you fixed your quotes. Yay.
>
> To answer your question, I would guess it has something to do with what
> the bash parser considers a "word". Brace expansion occurs when an
> unquoted part of a word contains a legitimate brace-expansion substring
> such as {a,b} or {a..c}.
>
> In your case, it looks like the parser considers the entire inside of the
> arithmetic command to be one "word", with a brace expansion inside it.
> So, the brace expansion is performed on that word, and you get
>
> ((1+ +0 2+ +0 3+ +0 ...))
>
> which you can confirm by re-running your command with set -x.
This is true. The contents of an arithmetic command are considered one
word, and are documented to be equivalent to `let' executed with the
contents as a single string.
So `(( ... ))' and `let "..."' are exactly equivalent.
The whitespace discrepancy might be caused by the fact that brace
expansion generates multiple words, which have to be rejoined into a
single string.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU c...@case.edu http://cnswww.cns.cwru.edu/~chet/
