Re: Howto declare a friend function to a nested class

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Sun, 30 Sep 2007 11:33:43 -0400
Message-ID:
<krqdncDv9utOXGLbnZ2dnUVZ_o2vnZ2d@comcast.com>
jdurancomas@gmail.com wrote:

[...]
Improved source code:

#include <iostream>

template <typename T>
class A
{
public:
 class B;
};

/* Example of template function */
template <typename T>
T add(T a, T b) {return a+b;}

template <typename T>
bool operator == (const typename A<T>::B &b1,
  const typename A<T>::B &b2);

template <typename T>
class A<T>::B
{
public:
 B(int a);

 friend bool operator == <> (const typename A<T>::B &b1,


Should be

    friend bool operator == <T> ...

      const typename A<T>::B &b2); // Line 29

private:
 int aa;
};

template <typename T>
A<T>::B::B(int a):
 aa(a)
{}

template <typename T>
bool operator == (const typename A<T>::B& b1,
  const typename A<T>::B& b2)
{
 return b1.aa == b2.aa;
}

class C {};

int main()
{
 A<C>::B b1(4), b2(5);

 bool res = (b1 == b2); // Line 57


There is no way for the compiler to determine that the template
operator should be used because from the expression 'b1 == b2'
the compiler cannot deduce the 'C' for the template -- the context
is not one of the deducible contexts.

 return 0;
}

The message error from compiler is:

nested.cpp: In instantiation of 'A<C>::B':
nested.cpp:55: instantiated from here
nested.cpp:29: error: template-id 'operator==<>' for 'bool
operator==(const A<C>::B&, const A<C>::B&)' does not match any
template declaration


That can be rectified by placing 'T' in the angle brackets, see
above.

nested.cpp: In function 'int main()':
nested.cpp:57: error: no match for 'operator==' in 'b1 == b2'


This cannot be corrected because the compiler canno deduce the 'T'
for the template operator==

I've googled a litle abut how declare template functions as friends,
and It looks likes that the current syntax is right.


Apparently it wasn't.

V
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