c++ - Declare a template function as friend -


i have global function this:

namespace x { namespace y { template <r, ...t> r foo(t&&... args) {     r r(args...);     return r; } } }

then in class a, want declare function foo friend of a. did:

class { template <r, ...t> friend r x::y::foo(t&&... args); a(int x, int y){} };

now when, call x::y::foo<a>(4, 5) fails compile error foo can not access private constructor of a. unable understand error, how declare foo friend of a correctly?

thanks in advance.

after fixing syntactic issues template parameters , parameter packs, seems work:

namespace x {     namespace y     {         template <typename r, typename ...t>         r foo(t&&... args)         {             r r(args...);             return r;         }     } }  class {     template <typename r, typename ...t>     friend r x::y::foo(t&&... args);     a(int x, int y){} };  int main() {     x::y::foo<a>(1, 2); }

here live example of above code compiling.


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