Re: stream operator of a class nested in a class template

From:

Victor Bazarov <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Mon, 20 Oct 2008 10:05:30 -0400

Message-ID:

<gdi37a$t4s$1@news.datemas.de>

Ares Lagae wrote:

How is an output stream operator of a class nested in a class template
defined? The code fragment below does not compile. Maybe it's just me,
but I don't see why it should not compile.

Best regards,

#include <iostream>

template <typename T>
struct foo
{
  struct bar {};
  bar bar_;
  const bar& get_bar() const { return bar_; }
};

template <typename Ch, typename Tr, typename T>
std::basic_ostream<Ch,Tr>& operator<<(std::basic_ostream<Ch,Tr>&
ostream, typename foo<T>::bar&)

^^^^^^^^^^^^^^^^^^^^^
Your argument is declared non-const here, but you pass a const ref.

Also, this is a non-deducible context. You will need to be explicit to
specify the 'T' template argument.

{
  return ostream;
}

int main()
{
  foo<int> f;
  const foo<int>::bar& b = f.get_bar();
  std::cout << b << "\n";
}

Comeau C/C++ 4.3.10.1 (Oct 6 2008 11:28:09) for
ONLINE_EVALUATION_BETA2
Copyright 1988-2008 Comeau Computing. All rights reserved.
MODE:strict errors C++ C++0x_extensions

"ComeauTest.c", line 21: error: no operator "<<" matches these
operands
            operand types are: std::ostream << const foo<int>::bar
    std::cout << b << "\n";
              ^

1 error detected in the compilation of "ComeauTest.c".

V
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