Re: Fix needed for communication with gpg-agent

[Kim F. Storm]

Chong Yidong <address@hidden> writes:

> address@hidden (Kim F. Storm) writes:
>
>> Noone has yet explained why using accept-process-output to wait just
>> for the pgp process and not process keyboard input while waiting is
>> not sufficient.
>>
>> If we need to enable SYNC_INPUT for that to work, and we don't want to
>> do that in general, we could make a negative "just-this-one" arg mean
>> "wait for single process to terminate with input blocked."
>>
>> I.e. a call like this:
>>
>>    (accept-process-output process nil nil -1)
>>
>> We would only have to do that on a terminal, as pinentry works fine on
>> X already.
>>
>> Any readon that wouldn't work??
>
> Have you tried it?  

No -- the version of gpg I have doesn't have the gpg-agent.
I haven't had time to upgrade.

>                     When I tried, gpg quits immediately, with or
> without SYNC_INPUT; I can't debug it because I don't understand how
> this idea is supposed to work in the first place.  IIUC, incoming tty
> input in stored by Emacs, never forwarded to the gpg process.

Ok, so you say that Emacs does monopolise the tty, reading all input (or
at least blocking anyone else from reading from the tty).

IIRC, someone else said that pinentry was able to read _something_ but
that it was very difficult to actually read a working password (because
Emacs would read some of the input).

So what's the truth here?


In any case, I support your proposal to document the current restrictions
for Emacs on terminals -- anything else seems too complex at this time.

After the release, I think we should work closely with the gpg team on
a long term solution for making Emacs 23 and pinentry co-operate also
on a terminal (maybe the multi-tty branch has some additional things
to consider too).

-- 
Kim F. Storm <address@hidden> http://www.cua.dk