Re: Clarification - Space removal in AE takes place before brace expansion

[Florian Mayer]

> OK.  This starts out as one word: "{1..10}'+' +0". Brace expansion
Ok, that's obviously a dumb mistake. And everything you wrote is right.

Just to clean up the mistake:
$ (( {1..10}'+' 0)) # spaces before { and 0

So, as you described, what bash actually does is it takes
$ (( {1..10}'+' 0))
goes into arithmetic expression context and performes brace
expansion by chosing {1..10} as the amble and "'+' 0" as postscript
So
$ "{1..10}'+' 0"
actually is seen as
$ "{1..10}'+ 0'" # postabmle = '+<space>0'
Then, it expands that to
'1+ 0 2+ 0 3+ 0 4+ 0 5+ 0 6+ 0 7+ 0 8+ 0 9+ 0 10+ 0'
and evaluates it arithmetically resulting in 55.

When I do
$ set -x
I get:
$ (( {1..10}'+' 0))
$ + ((  1+ 0  2+ 0  3+ 0  4+ 0  5+ 0  6+ 0  7+ 0  8+ 0  9+ 0  10+ 0 ))
$ bash: ((: 1+ 0  2+ 0  3+ 0  4+ 0  5+ 0  6+ 0  7+ 0  8+ 0  9+ 0 10+ 0: 
syntax error in expression (error token is "2+ 0  3+ 0  4+ 0 5+ 0  6+ 0  
7+ 0  8+ 0  9+ 0  10+ 0")
Which shows the exact same pattern.

> The whitespace discrepancy might be caused by the fact that brace
> expansion generates multiple words, which have to be rejoined into a
> single string.
But doesn't this case show that bash choses the preamble and postable
for the brace expansion wrongly?

> word (the space between the `+ +') is preserved.

Why is it preserved?
Were I to do
$ echo {1..10}'+' 0
I'd get a completely other set of words: '1+ 2+ 3+ 4+ 5+ 6+ 7+ 8+ 9+ 10+ 0'

I don't know how arithmetic expansion works, but
why can't we just brace expand everything in between (( ))
as if we'h read it verbatimly from the commandline?