On Sat, 16 Apr 2011 14:59:34 +0200, Pieter Praet <pieter@praet.org> wrote:
> On Thu, 14 Apr 2011 10:23:46 +0200, Florian Friesdorf <flo@chaoflow.net> wrote:
> > 
> > With 60k messages and 12k tagged as sent:
> > 
> > $ time notmuch tag +sent -- from:flo@chaoflow.net
> > 
> > real	0m8.561s
> > user	0m8.069s
> > sys	0m0.212s
> > 
> > $ time notmuch tag +sent -- from:flo@chaoflow.net and not tag:sent
> > 
> > real	0m0.043s
> > user	0m0.036s
> > sys	0m0.006s
> > 
> > 
> > This could be made implicit:
> > 
> > notmuch tag +A +B -- <filter>
> > -->
> > notmuch tag +A +B -- <filter> and not \(tag:A and tag:B\)
> > 
> > Apply command, if one of the tags is not set.
> > 
> > 
> > notmuch tag -C -D -- <filter>
> > -->
> > notmuch tag -C -D -- <filter> and \(tag:C or tag:D\)
> > 
> > Apply command, if one of the tags is set.
> > 
> > 
> > notmuch tag +A +B -C -D -- <filter>
> > -->
> > notmuch tag +A +B -C -D -- <filter> and \(not tag:A or not tag:B\ or tag:C or tag:D\)

The second '\' after 'B' is not supposed to be there.

> > In order to enforce tagging and disable the filter there could be a
> > flag.
> > 
> > I lack the knowledge/time to implement it, but I think it's at least
> > worth documenting it.
> 
> Most of us already do this explicitly in our tagging scripts, so no harm
> in making it standard behaviour, I guess.
> 
> Though to keep the implementation nice & clean, I'd advise against the
> use of parens: no need for escape chars, no messing with De Morgan's
> law, simply map the tag operations to their inverse in conjunctively
> joined filters:
> 
> notmuch tag +A +B -- <filter> and not tag:A or not tag:B
> notmuch tag -C -D -- <filter> and tag:C or tag:D
> notmuch tag +A +B -C -D -- <filter> and not tag:A or not tag:B or tag:C or tag:D

I am not sure whether I understand what you mean.

The parens are already supported by notmuch and also needed. Your second
line for example would remove C and D, if D is set, independently of
<filter>. Without parens, based on `and` taking precedence over `or`:

F and (C or D) = F and C or F and D.

-- 
Florian Friesdorf <flo@chaoflow.net>
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