pg.diff between two identical objects about pyglove HOT 5 CLOSED

Good question. It might seems confusing to return the same object itself, the mental model was to simulate a diff tool, which has the left side and the right side - when both sides are the same, they appear to be the same content, which is the object itself.

I think a better treatment is that when pg.diff(A(x='1'), A(x='1')) (mode='diff'), it should return None, and when the mode='same' or 'both', it returns the object itself. Does this make sense? I will be sending out a PR for this.

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I would prefer to be consistent with non-identical case.
For example, it make sense to me to see A(IDENTICAL) instead of None because the later case need additional handling on None. Also, if we have a simple api to just get a bool value for whether two objects are the same or not along with this diff, that would also be helpful (I guess we already have one?) :)

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It took me some revisit to realize that current behavior is actually correct.

The pg.diff(A(x='1', A(x='1')) returns pg.Diff(A(x='1'), A(x='1')) ('D' is capitalized) instead of A(x='1'), though str(pg.Diff) uses the str form of the value if left and right are identical. You can simply apply a bool operator on pg.Diff (e.g. bool(pg.Diff(...)) or use predicate like if not pg.diff(...).

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Okay, I tuned the behavior a little bit - now pg.diff(A(x='1'), A(x='1')) (mode='diff' by default) will return pg.Diff(left=pg.Diff.MISSING, right=pg.Diff.MISSING) whose representation or string is 'No diff'. For pg.diff(A(x='1', A(x='1'), mode='same'), it will return pg.Diff(left=A(x='1'), right=A(x='1')), which will print as A(x='1').

Please see Commit# e31fa99 for details.

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This makes a lot of sense to me, thanks for your quick fix!

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