No valid OpenPGP data found.
JanuszA.Urbanowicz
JanuszA.Urbanowicz
Wed Aug 29 18:59:01 2001
David Shaw wrote/napisa=B3[a]/schrieb:
> On Wed, Aug 29, 2001 at 09:59:25AM +0200, Werner Koch wrote:
> > On Wed, 29 Aug 2001 02:02:18 +0200, Marc Mutz said:
> >=20
> > > Shouldn't gnupg be more generous in what it accepts here? Eg. rfc2045=
=20
> > > explicitly requires MUA's to simply ignore any non-base64 character i=
n=20
> >=20
> > rfc2045 says in the section on ascii armor:
> >=20
> > starting five dashes, and following the ending five dashes. The
> > header lines, therefore, MUST start at the beginning of a line, and
> > MUST NOT have text following them on the same line. These line
>=20
> Don't you think this could be an example of "be conservative in what
> you generate, and liberal in what you accept"? Generate perfect
> RFC-compliant messages, sure, but maybe GnuPG should accept messages
> that are somewhat broken (especially in this case, which is a
> difference of whitespace only).
RFC 2440, page 46:
6.4. Decoding Radix-64
Any characters outside of the base64 alphabet are ignored in Radix-64
data. Decoding software must ignore all line breaks or other
characters not found in the table above.
In Radix-64 data, characters other than those in the table, line
breaks, and other white space probably indicate a transmission error,
about which a warning message or even a message rejection might be
appropriate under some circumstances.
Because it is used only for padding at the end of the data, the
occurrence of any "=3D" characters may be taken as evidence that the
end of the data has been reached (without truncation in transit). No
such assurance is possible, however, when the number of octets
transmitted was a multiple of three and no "=3D" characters are
present.
so, end-of-line-spaces are PEFECTLY legal.
Alex
--=20
C _-=3D-_ H| Janusz A. Urbanowicz | ALEX3-RIPE | SF-F Framling | | =
* =09
; (_O : +-------------------------------------------------------------+ --=
+~|=09
! &~) ? | P=B3yn=B1=E6 chc=EA na Wsch=F3d, za Suez, gdzie jest dobrem ka=
=BFde z=B3o | l_|/=09
A ~-=3D-~ O| Gdzie przykaza=F1 brak dziesi=EAciu, a pi=E6 mo=BFna a=BF po d=
no; | | =20