Re: How is the ^{sha256} peel syntax supposed to work?

[Jonathan Nieder <jrnieder@gmail.com>]

Hi,

Ævar Arnfjörð Bjarmason wrote:
> On Wed, Aug 29 2018, Jonathan Nieder wrote:

>> what objects would you expect the following to refer to?
>>
>>   abcdabcd^{sha1}
>>   abcdabcd^{sha256}
>>   ef01ef01^{sha1}
>>   ef01ef01^{sha256}
>
> I still can't really make any sense of why anyone would even want #2 as
> described above, but for this third case I think we should do this:
>
>     abcdabcd^{sha1}   = abcdabcdabcdabcdabcdabcdabcdabcdabcdabcd
>     abcdabcd^{sha256} = ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01
>     ef01ef01^{sha1}   = ef01ef01ef01ef01ef01ef01ef01ef01ef01ef01
>     ef01ef01^{sha256} = abcdabcdabcdabcdabcdabcdabcdabcdabcdabcd...
>
> I.e. a really useful thing about this peel syntax is that it's
> forgiving, and will try to optimistically look up what you want.

Sorry, I'm still not understanding.

I am not attached to any particular syntax, but what I really want is
the following:

	Someone who only uses SHA-256 sent me the commit id
	abcdabcdabcdabcdabcdabcdabcdabcdabcdabcd... out of band.
	Show me that commit.

	I don't care what object id you show me when you show that
	commit.  If I pass --output-format=sha1, then that means I
	care, and show me the SHA-1.

In other words, I want the input format and output format completely
decoupled.  If I pass ^{sha1}, I am indicating the input format.  To
specify the output format, I'd use --output-format instead.

That lets me mix both hash functions in my input:

	git --output-format=sha256 diff abcdabcd^{sha1} abcdabcd^{sha256}

I learned about these two commits out of band from different users,
one who only uses SHA-1 and the other who only uses SHA-256.

In other words:

[...]
> Similarly, I think it would be very useful if we just make this work:
>
>     git rev-parse $some_hash^{sha256}^{commit}
>
> And not care whether $some_hash is SHA-1 or SHA-256, if it's the former
> we'd consult the SHA-1 <-> SHA-256 lookup table and go from there, and
> always return a useful value.

The opposite of this. :)

Thanks,
Jonathan