I don't know what this c++ method means!

This is a discussion on I don't know what this c++ method means! within the C++ Programming forums, part of the General Programming Boards category; void Print(ostream&) const this constant member function will display data of the objects through the ostream object in the ...

  1. #1
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    I don't know what this c++ method means!

    void Print(ostream&) const this constant member function will display data of the objects through the ostream object in the argument.

    This is something I have to program.
    I guess I am suppose to take in an ostream object such as
    "cout"

    but i have no clue what to write
    Please help

  2. #2
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    Unhappy

    nobody knows?

  3. #3
    Registered User major_small's Avatar
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    your not gonna get an answer right away... this isn't a chat... when somebody sees the question and knows the answer, they'll post it...
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  4. #4
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    I think i figured it out

    I can take an object like

    obj.Print(cout).

    now my method can be


    cout<<(this.value);



    I think that's all there is to it?!?

  5. #5
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    This is kind of like overloading the >> ostream operator.
    Code:
    void ClassName::Print(&ostream os){
        os >> classvariable;
        os >> anotherprivatevariable;
    }
    .
    .
    Print should be public.
    .
    .
    ClassName.Print(cout);
    I think this is what you would do.

  6. #6
    Senior Member joshdick's Avatar
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    If that function is a member function of a class, then you should probably post something about the class, so that the rest of us know what that means.
    FAQ

    "The computer programmer is a creator of universes for which he alone is responsible. Universes of virtually unlimited complexity can be created in the form of computer programs." -- Joseph Weizenbaum.

    "If you cannot grok the overall structure of a program while taking a shower, you are not ready to code it." -- Richard Pattis.

  7. #7
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    curlious
    why did you use the >>operator?

    I read the api for cout and thought i had to use

    <<(const & primitive_type);

    ??

  8. #8
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    opps my bad you are correct it is the << operator but using
    cout << this.value will not work if value is private.
    classname.Print(cout); will work if you define the function like I said.

    Code:
    you can overload the << operator like this
    
    ostream & operator<<(ostream & os, const ClassName & object){
       os<< classvar;
       return os;
    }
    
    you should make the prototype a friend function in your class declaration
    
    then you can do this in main
    cout << classobject;

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