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Sep 10 at 9:08 review Close votes
Sep 14 at 0:02
Sep 10 at 8:47 comment added Donald Duck This question is similar to: Is there a shorter way to forward declare a class in a namespace?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.
S Jul 22, 2021 at 7:38 history edited Dada CC BY-SA 4.0
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Jul 22, 2021 at 7:32 review Suggested edits
S Jul 22, 2021 at 7:38
Apr 9, 2020 at 3:12 answer added user2193043 timeline score: 40
Oct 31, 2018 at 9:32 answer added Campa timeline score: 5
S May 14, 2014 at 21:28 history suggested arserbin3 CC BY-SA 3.0
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May 14, 2014 at 21:26 review Suggested edits
S May 14, 2014 at 21:28
Oct 1, 2013 at 15:48 audit Suggested edits
Oct 1, 2013 at 15:48
Sep 25, 2013 at 10:58 vote accept Angus Comber
Sep 25, 2013 at 10:58 history edited Angus Comber CC BY-SA 3.0
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Sep 25, 2013 at 10:38 history edited Angus Comber CC BY-SA 3.0
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Sep 25, 2013 at 10:35 comment added Avi Perel this is not just a namespace issue. consumer.cpp is aware that there is a class type a, but it is not aware of the specifics. You try to invoke the method a::talk(), which the compiler knows nothing about. You still need to #include "a.hpp" from consumer.cpp so the compiler is aware of the classes full interface. This #include will be internal to the .cpp hence will not be "spread around" through consumer.hpp.
Sep 25, 2013 at 10:14 history edited Angus Comber CC BY-SA 3.0
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Sep 25, 2013 at 10:01 history edited Angus Comber CC BY-SA 3.0
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Sep 25, 2013 at 9:59 comment added Benjamin Bannier Don't use names starting with two underscores or an underscore and a capital letter for macros -- these are reserved. stackoverflow.com/questions/17307540/…
Sep 25, 2013 at 9:56 answer added billz timeline score: 137
Sep 25, 2013 at 9:55 history asked Angus Comber CC BY-SA 3.0