Inheritance downcast query

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  • jayesah@gmail.com
    #1

    Inheritance downcast query

    Hi All,

    I have following code. Anyone can please tell me how
    object of type class A is converted to Class B ?

    #include <iostream.h>
    class A
    {
    public:
    void operator=(int i){a=i;}

    A operator+(A& obj)
    {
    A nn;
    nn.a = a + obj.a;
    cout<<"A: operator + is called\n";
    return nn;

    }

    int a;
    };

    class B : public A
    {
    using A::operator=;

    public:
    B()
    { cout<<"b's simple called\n"; };
    int b;
    };

    int main(void)
    {
    B bb1,bb2;
    bb1=3;
    bb2=4;

    B bb3;
    bb3=bb1+bb2; /* Here, Who converted type Class B to Class A */

    cout << bb3.a;

    return 0;
    }

    Tags: None
    • Alf P. Steinbach
      #2
      Re: Inheritance downcast query

      * jayesah@gmail.c om:
      >
      I have following code. Anyone can please tell me how
      object of type class A is converted to Class B ?
      Yes, but first, about the code.

      #include <iostream.h>
      This is not a standard header. E.g., Visual C++ 7.x and higher doesn't
      have this header. Use standard <iostreamplus <ostream(and perhaps
      <istreamif input is required).

      class A
      {
      public:
      void operator=(int i){a=i;}
      >
      A operator+(A& obj)
      That argument should be "A const& obj" unless you're intent on changing
      the actual argument object.

      As it is you can't call "+" with a constant or temporary right hand side.

      {
      A nn;
      nn.a = a + obj.a;
      cout<<"A: operator + is called\n";
      return nn;
      >
      }
      >
      int a;
      };
      >
      class B : public A
      {
      using A::operator=;
      Are you sure you want this assignment operator private?

      public:
      B()
      { cout<<"b's simple called\n"; };
      You mean, "default constructor".

      int b;
      };
      >
      int main(void)
      {
      B bb1,bb2;
      bb1=3;
      bb2=4;
      >
      B bb3;
      bb3=bb1+bb2; /* Here, Who converted type Class B to Class A */
      First, there's no conversion from B to A, and this should not compile
      (have you actually tried the code?).

      Second, if you make the carte blanche "using" statement public it should
      compile, because the auto-generated A::operator=( A const& ) is then
      made available in B and invoked. To avoid that you just need to be a
      bit more specific.

      E.g., in class B, "void operator=( int i ){ A::operator=( i ); }"; or
      better, use a conventional member function instead of an operator.

      cout << bb3.a;
      >
      return 0;
      }
      >

      --
      A: Because it messes up the order in which people normally read text.
      Q: Why is it such a bad thing?
      A: Top-posting.
      Q: What is the most annoying thing on usenet and in e-mail?

        Comment

        • Naresh
          #3
          Re: Inheritance downcast query


          Alf P. Steinbach wrote:
          * jayesah@gmail.c om:

          I have following code. Anyone can please tell me how
          object of type class A is converted to Class B ?
          >
          Yes, but first, about the code.
          >
          >
          #include <iostream.h>
          >
          This is not a standard header. E.g., Visual C++ 7.x and higher doesn't
          have this header. Use standard <iostreamplus <ostream(and perhaps
          <istreamif input is required).
          >
          >
          class A
          {
          public:
          void operator=(int i){a=i;}

          A operator+(A& obj)
          >
          That argument should be "A const& obj" unless you're intent on changing
          the actual argument object.
          >
          As it is you can't call "+" with a constant or temporary right hand side.
          >
          >
          {
          A nn;
          nn.a = a + obj.a;
          cout<<"A: operator + is called\n";
          return nn;

          }

          int a;
          };

          class B : public A
          {
          using A::operator=;
          >
          Are you sure you want this assignment operator private?
          >
          >
          public:
          B()
          { cout<<"b's simple called\n"; };
          >
          You mean, "default constructor".
          >
          >
          int b;
          };

          int main(void)
          {
          B bb1,bb2;
          bb1=3;
          bb2=4;

          B bb3;
          bb3=bb1+bb2; /* Here, Who converted type Class B to Class A */
          >
          First, there's no conversion from B to A, and this should not compile
          (have you actually tried the code?).
          >
          Second, if you make the carte blanche "using" statement public it should
          compile, because the auto-generated A::operator=( A const& ) is then
          made available in B and invoked. To avoid that you just need to be a
          bit more specific.
          >
          E.g., in class B, "void operator=( int i ){ A::operator=( i ); }"; or
          better, use a conventional member function instead of an operator.
          >
          >
          cout << bb3.a;

          return 0;
          }
          >
          >
          --
          A: Because it messes up the order in which people normally read text.
          Q: Why is it such a bad thing?
          A: Top-posting.
          Q: What is the most annoying thing on usenet and in e-mail?

          its a well compiled code with gcc 3.3.3 on linux.
          It also gives expected output i..e 7

            Comment

            • Naresh
              #4
              Re: Inheritance downcast query


              Alf P. Steinbach wrote:
              * jayesah@gmail.c om:

              I have following code. Anyone can please tell me how
              object of type class A is converted to Class B ?
              >
              Yes, but first, about the code.
              >
              >
              #include <iostream.h>
              >
              This is not a standard header. E.g., Visual C++ 7.x and higher doesn't
              have this header. Use standard <iostreamplus <ostream(and perhaps
              <istreamif input is required).
              >
              >
              class A
              {
              public:
              void operator=(int i){a=i;}

              A operator+(A& obj)
              >
              That argument should be "A const& obj" unless you're intent on changing
              the actual argument object.
              >
              As it is you can't call "+" with a constant or temporary right hand side.
              >
              >
              {
              A nn;
              nn.a = a + obj.a;
              cout<<"A: operator + is called\n";
              return nn;

              }

              int a;
              };

              class B : public A
              {
              using A::operator=;
              >
              Are you sure you want this assignment operator private?
              >
              >
              public:
              B()
              { cout<<"b's simple called\n"; };
              >
              You mean, "default constructor".
              >
              >
              int b;
              };

              int main(void)
              {
              B bb1,bb2;
              bb1=3;
              bb2=4;

              B bb3;
              bb3=bb1+bb2; /* Here, Who converted type Class B to Class A */
              >
              First, there's no conversion from B to A, and this should not compile
              (have you actually tried the code?).
              >
              Second, if you make the carte blanche "using" statement public it should
              compile, because the auto-generated A::operator=( A const& ) is then
              made available in B and invoked. To avoid that you just need to be a
              bit more specific.
              >
              E.g., in class B, "void operator=( int i ){ A::operator=( i ); }"; or
              better, use a conventional member function instead of an operator.
              >
              >
              cout << bb3.a;

              return 0;
              }
              >
              >
              --
              A: Because it messes up the order in which people normally read text.
              Q: Why is it such a bad thing?
              A: Top-posting.
              Q: What is the most annoying thing on usenet and in e-mail?

              its a well compiled code with gcc 3.3.3 on linux.
              It also gives expected output i..e 7

                Comment

                • Alf P. Steinbach
                  #5
                  Re: Inheritance downcast query

                  * Naresh:
                  >
                  its a well compiled code with gcc 3.3.3 on linux.
                  It also gives expected output i..e 7
                  It shouldn't compile, so if you actually copied and pasted the code that
                  compiled for you, then get a better compiler.

                  Btw., please quote what you're responding to but /no more/. Like
                  Einstein's "as simple as possible, but no simpler". Also, please don't
                  quote signatures, and please don't post the same message two or more times.

                  HTH. & TIA.


                  --
                  A: Because it messes up the order in which people normally read text.
                  Q: Why is it such a bad thing?
                  A: Top-posting.
                  Q: What is the most annoying thing on usenet and in e-mail?

                    Comment

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