Which way does the standard mandate to forward declare types that belongs to
namespace std. In particular I want to declare the streaming operator <<
for some class in a headerfile, without having to include <iostream>.

Perhaps <iosfwd> can solve this particular problem but what about a more
general case where the type is not declared in that header file.

Example:
namespace std { class ostream; }
// Is the line above legal according to the standard?
// If not, what are the alternatives?

class Foo
{
public:
  friend std::ostream& operator<< (std::ostream&, const Foo&);
  //...
};

/Niklas Mellin


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<nimel@my-dejanews.com> wrote:
>
>Which way does the standard mandate to forward declare types that belongs to
>namespace std. In particular I want to declare the streaming operator <<
>for some class in a headerfile, without having to include <iostream>.

It is entirely forbidden.  If you declare a name in std:: without
first including the header that defines it, your program is broken.
The compiler is not required to report the error.

There is only one way to get a standard name into your program, and
that is via #include.  This is not an idle restriction; if you violate
it your code may break without warning.

>Perhaps <iosfwd> can solve this particular problem but what about a more
>general case where the type is not declared in that header file.

Then you can include <ios>, or <ostream>, or <istream>, instead.

--
Nathan Myers
ncm@nospam.cantrip.org  http://www.cantrip.org/



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nimel@my-dejanews.com wrote:
>
> Which way does the standard mandate to forward declare types that belongs to
> namespace std. In particular I want to declare the streaming operator <<
> for some class in a headerfile, without having to include <iostream>.
>
> Perhaps <iosfwd> can solve this particular problem but what about a more
> general case where the type is not declared in that header file.
>
> Example:
> namespace std { class ostream; }

There is no "class ostream;" in namespace std {}.
<iosfwd> defines std::ostream as
 namespace std {
  typedef basic_ostream<char> ostream;
 }
> // Is the line above legal according to the standard?
Maybe ... probably not.
<quote>
 17.3.1.2 Requrements [lib.structure.requirements]
 The library can be extended by a C++ program. Each clause as
 applicable describes the requirements that such extensions must meet.
</quote>
You didn't want to extend the library, did you ?

 --ALfred

> // If not, what are the alternatives?
>
> class Foo
> {
> public:
>   friend std::ostream& operator<< (std::ostream&, const Foo&);
>   //...
> };



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