friend oper<<

[vineoff]

Why do I have to declare my oper<< as friend in my class as follows:

class A {
public:
friend std::ostream& operator<<(std: :ostream& out, const A& a)
{ return out << "foo"; }
};

[marco]

vineoff ha scritto:

Why do I have to declare my oper<< as friend in my class as follows:

class A {
public:
friend std::ostream& operator<<(std: :ostream& out, const A& a)
{ return out << "foo"; }
};

Why did you do? It is necessary only if you want to have access to
non-public members of class A within operator <<.

In this case you can write
class A {};
std::ostream& operator<<(std: :ostream& out, const A& a)
{ return out << "foo"; }

[Rolf Magnus]

vineoff wrote:

Why do I have to declare my oper<< as friend in my class as follows:

class A {
public:
friend std::ostream& operator<<(std: :ostream& out, const A& a)
{ return out << "foo"; }
};

No reason, i.e. you don't have to.

[vineoff]

But can you show me an example without friend keyword. If I just remove
it now, it won't compile

[xian]

vineoff wrote:

But can you show me an example without friend keyword. If I just remove
it now, it won't compile

like this :)

#include<iostre am>

class A {
public:
int val;
};
std::ostream& operator<<(std: :ostream& out, const A& a)
{
return out << a.val;
}

int main()
{
A a;
std::cout << a;
}

And Scott Meyers said: Don't let operator << or operator >> be the
members. And if you want to access the private members, let it be a
friend.

:)

[Rolf Magnus]

vineoff wrote:

But can you show me an example without friend keyword. If I just remove
it now, it won't compile

The operator must be a non-member, which it is automatically if you declare
it as friend. So you have to write:

class A {
public:
};

std::ostream& operator<<(std: :ostream& out, const A& a)
{ return out << "foo"; }

int main()
{
}

[vineoff]

Why operator== i.e. can be declared as member function but operator<<
can't?

[Thomas Tutone]

vineoff wrote:

Why operator== i.e. can be declared as member function but operator<<
can't?

This is an FAQ:

http://www.parashift.com/c++-faq-lit....html#faq-15.8

It's not that you can't declare operator<<() as a member function, it's
that it wouldn't do what you seem to think it would.

In short, to declare operator<<() as a member function, you would have
to modify the declaration and definition of class std::ostream, since
the left-hand-side parameter is of type std::ostream and the
right-hand-side parameter is the object you want to send to the
ostream. And of course, you are not permitted to modify std::ostream
yourself.

If you, on the other hand, made operator<<() a member of your class
Foo, then you would have to put Foo on the left-hand-side and the
std::ostream on the right-hand-side when you used the operator, which
would look very strange.

Best regards,

Tom

[Thomas Tutone]

vineoff wrote:

Why operator== i.e. can be declared as member function but operator<<
can't?

This is an FAQ:

http://www.parashift.com/c++-faq-lit....html#faq-15.8

It's not that you can't declare operator<<() as a member function, it's
that it wouldn't do what you seem to think it would.

In short, to declare operator<<() as a member function, you would have
to modify the declaration and definition of class std::ostream, since
the left-hand-side parameter is of type std::ostream and the
right-hand-side parameter is the object you want to send to the
ostream. And of course, you are not permitted to modify std::ostream
yourself.

If you, on the other hand, made operator<<() a member of your class
Foo, then you would have to put Foo on the left-hand-side and the
std::ostream on the right-hand-side when you used the operator, which
would look very strange.

Best regards,

Tom

[Thomas Tutone]

Thomas Tutone wrote:

[Quote of repetitive post snipped ...]

Well, I posted this ONCE at around 1:30pm New York time via Google
Groups, and it has appeared TWICE on usenet - once at 2:37pm, and again
at 3:01pm.

Goodness only knows what is going on at Google...

Best regards,

Tom