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Home/ Questions/Q 704327
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T03:55:46+00:00 2026-05-14T03:55:46+00:00

In code: struct Rep { const char* my_data_; Rep* my_left_; Rep* my_right_; Rep(const char*);

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In code:

struct Rep
    {
        const char* my_data_;
        Rep* my_left_;
        Rep* my_right_;
        Rep(const char*);
    };

typedef Rep& list;

ostream& operator<<(ostream& out, const list& a_list)
    {
        int count = 0;
        list tmp = a_list;//----->HERE I'M CREATING A LOCAL COPY
        for (;tmp.my_right_;tmp = *tmp.my_right_)
        {
            out << "Object no: " << ++count << " has name: " << tmp.my_data_;
            //tmp = *tmp.my_right_;
        }
        return out;//------>HERE a_list is changed
    }

I’ve thought that if I’ll create local copy to a_list object I’ll be operating on completely separate object. Why isn’t so?

Thanks.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T03:55:47+00:00Added an answer on May 14, 2026 at 3:55 am

    I assume list is meant to be the same as Rep.

    You are only copying the the pointer to the first node in the list. You are not copying the data, nor the rest of the nodes of the list. You are doing a shallow copy of the first node of the list. If you would also copy the objects themselves it would be deep copy.

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