Re: Question on rerere

Junio C Hamano [off-list ref] writes:

Johannes Schindelin [off-list ref] writes:

quoted

IIUC then each conflict hunk is handled _separately_ like this: the 
lexicographically smaller between the two file sections is displayed 
first, regardless if a previous hunk had a different order. Not that it 
matters most of the time, but isn't this dangerous?

You are probably right.  Probably the right thing would be to
use the first hunk to determine the flipping order and stick to
that for the rest.

Not that I've seen problems with the current behaviour, though.

Well, come to think of it, I think the current behaviour makes
more sense.

Suppose you start from an original file "OO".  You have two
branches that change it to "AO" and "BO", and another pair of
branches that change it to "OC" and "OD".  Let's call these
branches A, B, C, and D.

By merging A and C, you will get "AC"; you can get "AD", "BC"
and "BD" the same way.

Now suppose you are on "AC" and merged "BD".  You would get
"<A=B><C=D>".

If you were on "BD" and merged "AC" you would get "<B=A><D=C>".
If you were on "AD" and merged "BC" you would get "<A=B><D=C>".

You got the idea?

Hi,

On Tue, 19 Dec 2006, Junio C Hamano wrote:

Junio C Hamano [off-list ref] writes:

quoted

Johannes Schindelin [off-list ref] writes:

quoted

IIUC then each conflict hunk is handled _separately_ like this: the 
lexicographically smaller between the two file sections is displayed 
first, regardless if a previous hunk had a different order. Not that it 
matters most of the time, but isn't this dangerous?

You are probably right.  Probably the right thing would be to
use the first hunk to determine the flipping order and stick to
that for the rest.

Not that I've seen problems with the current behaviour, though.

Well, come to think of it, I think the current behaviour makes
more sense.

Suppose you start from an original file "OO".  You have two
branches that change it to "AO" and "BO", and another pair of
branches that change it to "OC" and "OD".  Let's call these
branches A, B, C, and D.

By merging A and C, you will get "AC"; you can get "AD", "BC"
and "BD" the same way.

Now suppose you are on "AC" and merged "BD".  You would get
"<A=B><C=D>".

If you were on "BD" and merged "AC" you would get "<B=A><D=C>".
If you were on "AD" and merged "BC" you would get "<A=B><D=C>".

You got the idea?

Yes. Thanks!

Ciao,
Dscho