Re: Test scalar refs (not values) in deeply nested structure

The Test::More documentation says for is_deeply:

is_deeply() compares the dereferenced values of references, the references themselves (except for their type) are ignored.

... so I'm not sure what you expect.

If you want to compare the identity of the references too, you will need to write your own comparison tool, but I guess that a simple shallow comparison already is enough for that case.

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Re^2: Test scalar refs (not values) in deeply nested structure by LanX (Saint) on Jul 23, 2020 at 19:58 UTC
unfortunately my data is far from being shallow. My pragmatic approach is to write a recursive diver which will replace the scalar values with the stringification of their refs. `$$ref = "$ref"` I'll feed this to `is_deeply()` then, seems to work so far. Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l] [select]
Re^3: Test scalar refs (not values) in deeply nested structure by Corion (Patriarch) on Jul 23, 2020 at 20:21 UTC
So you have a nested data structure, but you want only scalar refs to be compared for identity, while you want to compare everything else structurally? If that's not it, maybe show examples and counterexamples of what you want your comparison routine to do.	[reply]
Re^4: Test scalar refs (not values) in deeply nested structure by LanX (Saint) on Jul 23, 2020 at 20:31 UTC
> only scalar refs to be compared for identity, while you want to compare everything else structurally? yes. > maybe show examples and counterexamples see here -> Re: Test scalar refs (not values) in deeply nested structure (workaround) Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply]


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