Re: printf and the $ modifier

[Chet Ramey]

On 12/19/14, 9:46 AM, Maarten Billemont wrote:
> man 3 printf describes the following:
> 
>      o   An optional field, consisting of a decimal digit string followed
> by a $, specifying the next argument to access.  If this field is not 
> provided,
>          the argument following the last argument accessed will be used. 
> Arguments are numbered starting at 1.  If unaccessed arguments in the format
>          string are interspersed with ones that are accessed the results
> will be indeterminate.
> 
> This is a useful feature, allowing you to decouple the order of your
> arguments with from the format.  This is useful for localization and a few
> other cases.  In this example, I'd like to use it to simplify a regex
> replace statement:
> 
> # s/(.*)foo(.*)/$2bar$1/
> [[ 123foo567 =~ (.*)foo(.*) ]] && printf '%2$sbar%1$s' "${BASH_REMATCH[@]:1}"
> 
> Is there a particular reason why bash's built-in printf does not support
> this format modifier?  Does bash re-implement printf or does it use the
> Standard C Library's printf?  (If the former; why?)

Several reasons:

1. Posix doesn't require it.

2. It doesn't mesh well with the argument-reuse and argument-
   marshaling-for-printf requirements, especially since that modifier
   has to be used for every argument or things don't work.

3. There's not actually a lot of demand to make it available, and few
   implementations go through the pain (even the standalone GNU printf).
   The only one I found after a quick non-exhaustive search is the ksh93
   builtin, which doesn't use the libc printf engine at all.

Chet

-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer
                 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU    chet@case.edu    http://cnswww.cns.cwru.edu/~chet/