Re: Create Global Function and Class

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 25 Aug 2010 01:37:53 -0700 (PDT)
Message-ID:
<359e4cee-5a0e-4ab5-b821-c4ad37b2fa5b@q22g2000yqm.googlegroups.com>
On Aug 24, 4:23 pm, Immortal Nephi <Immortal_Ne...@hotmail.com> wrote:

        I need to follow best practice guideline how to design my code
correctly. I have to decide which either global function or class can
be chosen.
        I will use global function if global function is available to all
file scopes or is in namespace scope. I need to avoid data
modification when necessary.
        I can write good global function.

int gFunc( int x, int y ) {
        // Do something
        // =85.

        return z;
}

        If I have to modify either x or y in the function parameter, I wi=

ll

put const in function parameter or create temporary variable.

int gFunc( const int x, const int y ) {


Note that the const above does not affect the function signature
(the client view of the function) in any way.

        // Do something
        // =85.

        return z;
}

// or=85

int gFunc( int x, int y ) {
        int temp_x = x;
        int temp_y = y;


Not sure what you're trying to accomplish with the temporaries.
You'll have to explain yourself more clearly.

        // Do something
        // =85.

        return z;
}

        If I want to put character buffer (like string) or class name in =

the

function parameter, I will need to put reference in it to avoid
overhead of copying each data size.


Maybe. It's a common convention, anyway.

Also, I will remember to put const before reference if
necessary.

class Object {
        int x;
        int y;
        int z;
}

Object gFunc( const Object& obj ) {
        Object temp_obj;

        // Do something
        // =85.

        return temp_obj;
}

        The return type requires to create temporary object before all da=

ta

members in class Object is copied and temp_obj is destroyed from the
stack.

        Tell me why you have reason to add reference to the return type.


Because you want to return a reference to an existing object,
rather than a new object. The classical example is
std::vector<>::operator[]. (If it is a search function which
can fail, you might prefer returning a pointer, with a null
pointer for failure.)

class Data {
        int x;
        int y;
        int z;
};

class Object {
        Data m_Data;

        Data &DoSomething() {
                // Do something
                return m_Data;
        }
};

        Notice DoSomething has reference return type. Do you need to use
reference return type? If yes, what is it used for?


It's used to allow the client access to the actual object.

        Maybe, you want to suggest your code to explain why you need
reference return type. Reference return type is dangerous unless you
use return =91this' pointer.


Reference return types are necessary for some things to work.

--
James Kanze

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