[ietf-dkim] Re: bh=; l=0;

[Paul Hoffman]

At 7:23 AM -0700 7/2/06, Michael Thomas wrote:
>Paul Hoffman wrote:
>
>>
>>
>>>#13 3.4.5, 3.5 and 3.7. "l=0" is allowed, but "bh=" is REQUIRED, 
>>>which is a bit
>>>of a contradiction.
>>
>>
>>"l=0;bh=;" seems valid.
>
>
>It doesn't seem valid to me, and it's certainly not the natural thing that
>an implementor would do which is to just take the value of SHA[1|256]_Final
>and compare it against the bh= value. Leaving it as is doesn't create an
>unnecessary new case.

Oh, wait; Michael is completely right here. If l=0, bh= contains the 
hash of a zero-length string, which has the length of a normal hash 
output. My apologies.

>>>And "l=" is not mentioned when saying how to calculate
>>>"bh=". I guess the right thing to do might be to add some mention 
>>>of "l=" when
>>>talking about calculating "bh=",
>>
>>
>>Agree.
>
>
>I don't see what the problem is: l= is the canonical byte count, and that's
>just as true with bh as is was before bh was invented.

The question is, what string do you hash when l= is given? I am 
interpreting the spec as that you has the string with the length 
given in l=; are you reading it as you hash the entire body 
regardless of the presence of l=?