I have template <int N> class A; and template <typename T> class B;. How do I declare B<A<N>> to be a friend for all N? template <int N> friend B<A<N>>; does not work.
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https://stackoverflow.com/questions/3292795/how-to-declare-a-templated-struct-class-as-a-friend – Mat Feb 18 '22 at 19:40
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2@Mat That looks different, as OP only wants a subset of specializations of `B`. – HolyBlackCat Feb 18 '22 at 19:52
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1A partial specialisation cannot be a friend. – n. 1.8e9-where's-my-share m. Feb 18 '22 at 20:34
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Sorry for the intrusion. You said, "How do I declare `B>` to be a friend for all `N`" - a friend to *what* exactly. `A
` ? The who-is-friending-who-and-where failure attempt in full-code would probably be a nice addition to this question, or at least make it clearer to people that don't necessarily sleep with the standard under their pillow. – WhozCraig Feb 18 '22 at 20:56 -
1@WhozCraig a friend to whatever class has a friend declaration I suppose. It doesn't really matter which class it is. You can try `class C { template
friend B>; };` (which doesn't work). – n. 1.8e9-where's-my-share m. Feb 18 '22 at 22:06 -
@n.1.8e9-where's-my-sharem. yeah, the broken syntax (missing `class/struct`) was what was confusing me. I get it now. And I agree with your first comment. – WhozCraig Feb 18 '22 at 22:31
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@WhozCraig You can add `class/struct` and it still doesn't work, that's kinda the problem. – n. 1.8e9-where's-my-share m. Feb 18 '22 at 22:36
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1Duplicate https://stackoverflow.com/questions/11041274 – cigien Feb 19 '22 at 00:07
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@n.1.8e9-where's-my-sharem. As I said, I get it now; thus the reason for the "I agree with your first comment" conclusion. – WhozCraig Feb 19 '22 at 02:57