friend oper<<

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  • vineoff
    #1

    friend oper<<

    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"; }
    };

    Tags: None
    • marco
      #2
      Re: friend oper&lt;&lt;

      vineoff ha scritto:[color=blue]
      > 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"; }
      > };
      >[/color]
      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"; }

        Comment

        • Rolf Magnus
          #3
          Re: friend oper&lt;&lt;

          vineoff wrote:
          [color=blue]
          > 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"; }
          > };[/color]

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

            Comment

            • vineoff
              #4
              Re: friend oper&lt;&lt;

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

                Comment

                • xian
                  #5
                  Re: friend oper&lt;&lt;


                  vineoff wrote:[color=blue]
                  > But can you show me an example without friend keyword. If I just remove
                  > it now, it won't compile[/color]

                  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.

                  :)

                    Comment

                    • Rolf Magnus
                      #6
                      Re: friend oper&lt;&lt;

                      vineoff wrote:
                      [color=blue]
                      > But can you show me an example without friend keyword. If I just remove
                      > it now, it won't compile[/color]

                      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()
                      {
                      }

                        Comment

                        • vineoff
                          #7
                          Re: friend oper&lt;&lt;

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

                            Comment

                            • Thomas Tutone
                              #8
                              Re: friend oper&lt;&lt;

                              vineoff wrote:[color=blue]
                              > Why operator== i.e. can be declared as member function but operator<<
                              > can't?[/color]

                              This is an FAQ:

                              Input/output via and , C++ FAQ
                              http://www.parashift.com/c++-faq-lite/input-output.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

                                Comment

                                • Thomas Tutone
                                  #9
                                  Re: friend oper&lt;&lt;

                                  vineoff wrote:[color=blue]
                                  > Why operator== i.e. can be declared as member function but operator<<
                                  > can't?[/color]

                                  This is an FAQ:

                                  Input/output via and , C++ FAQ
                                  http://www.parashift.com/c++-faq-lite/input-output.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

                                    Comment

                                    • Thomas Tutone
                                      #10
                                      Re: friend oper&lt;&lt;

                                      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

                                        Comment

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