Re: flag z won't cause forced joining?

[Han Pingtian <hanpt@linux.vnet.ibm.com>]

On Fri, Oct 19, 2012 at 08:38:18AM -0700, Bart Schaefer wrote:
> On Oct 19,  4:03pm, Han Pingtian wrote:
> }
> } I just found that it looks like the z flag won't cause "forced joining"
> } which stated in rules 10
> 
> Yes, this is a case of the documentation being wrong; it was recently
> corrected.  Rule 10 now says:
> 
> 10. _Forced joining_
>      If the `(j)' flag is present, or no `(j)' flag is present but the
>      string is to be split as given by rule 11., and joining did not
>      take place at step 5., any words in the value are joined together
>      using the given string or the first character of $IFS if none.
>      Note that the `(F)' flag implicitly supplies a string for joining
>      in this manner.
> 
> 11. _Simple word splitting_
>      If one of the `(s)' or `(f)' flags are present, or the `='
>      specifier was present (e.g. ${=VAR}), the word is split on
>      occurrences of the specified string, or (for = with neither of the
>      two flags present) any of the characters in $IFS.
> 
>      If no `(s)', `(f)' or `=' was given, but the word is not quoted
>      and the option SH_WORD_SPLIT is set, the word is split on
>      occurrences of any of the characters in $IFS.  Note this step, too,
>      takes place at all levels of a nested substitution.
> 
> Note that the reference to rule 17 [the (z) flag] has been removed from
> rule 10 (and rule 11 is new).

Thanks a lot.