c++ - Explicit friend specialization of template function in class template -

i'm facing situation arise 2 questions how adl , template function specialization works under circumstances.

1. why can't use cout inside friend specialization definition? in scenario i'm getting link error stating basic_stream undefined, if switch call cat compilation proceed.

   template<class t> void func1(t&){    ... }  void cat(){ cout << "foo.func1" <<endl;  }  namespace first{     template<class r>     struct foo{        friend void func1<>(foo<int>&){           cout << "foo.func1" <<endl;   // cat();        }             };        }  foo<int> f; func1(f); 

2. why adl doesn't apply when change specialization refer template class param? if i'm not wrong adl mechanic, in order resolve right version of func1 call in 3 (e.g. global(1), or friend defined one(2)) collects possible matches , selects concrete one. and, think concrete 1 version of func1 in 2, because instead of specifying function param dependent of template param, specify concrete type foo func1 in 1 doesn't, adl select func1 in 1 anyway. why ?

   template<class t> void func1(t&){     // 1     ... }  namespace first{     template<class r>     struct foo{        friend void func1<>(foo<r>&){        // 2           cout << "foo.func1" <<endl;        }             };       }   foo<int> f; func1(f);       // 3 

template parameters unrelated friend declarations. you'll need carry them disambiguated in thefriend declaration:

template<class r> struct foo{    template<typename u>    friend void func1<u>(foo<u>&){       cout << "foo.func1" <<endl;   // cat();    }         };       

also case should decide, if want put friend definition inlined above, or provide declaration:

template<class r> struct foo{    template<typename u>    friend void ::func1<u>(foo<u>&); };       

the latter should match friend template function in global namespace explicitly, , specialization can made necessary:

template<> void func1(int&){    // ... }  template<> void func1(std::string&){    // ... }  // a.s.o.