Re: why must non-standard $IFS members be treated so differently ?

[Dan Douglas]

On Sunday, July 29, 2012 03:23:29 PM Jason Vas Dias wrote:
> echo $(count_args 1 2 3\ 4)

I should also have mentioned that I couldn't reproduce this case. You should 
be getting 4 here in your example, not 3. I have the same Bash version. Are 
you sure you were echoing address@hidden ' and not address@hidden', and also 
that you did 
not set IFS=: for count_args? If you use exactly the function you sent  with 
the default IFS then you should get 4 here.
-- 
Dan Douglas

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