Re: C++ Union compilation errors

From:
Greg Herlihy <greghe@mac.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sat, 16 Aug 2008 21:22:45 CST
Message-ID:
<a596ee06-2c13-4425-b63e-dd3a0f4f3d7c@t1g2000pra.googlegroups.com>
On Aug 14, 3:39 pm, Sujal <sujal.sh...@gmail.com> wrote:

In below program, I'm getting below compilation errors

error C2621: member '_tag_nodes_::node_' of union '_tag_nodes_' has
copy constructor
error C2621: member '_tag_nodes_::linkednode_' of union '_tag_nodes_'
has copy constructor

Can anybody help me resolve this or any workaround for this?


The error message is pretty much self-explanatory - members of a union
may not have copy constructors (or constructors of any kind, for that
matter). The problem is that the the compiler would not know which
member copy constructor to call whenever the union itself was copied.
(In other words, the compiler does not keep track which member of a
union happens to be "active", that is, which member currently holds a
value).

C++ programs tend to avoid unions altogether - preferring instead to
use class hierarchies (often with polymorphic behavior). For example,
I would start with basic class hierarchy along these lines:

     #include <string>

     using std::string;

     struct node_info
     {
                         node_info();
                         node_info(const node_info&);
         node_info& operator=(const node_info&);

         node_info * next_node;
         string name;
         string type;
     };

     struct path_node : public node_info
     {
                     path_node();
                     path_node(const path_node&);
         path_node& operator=(const path_node&);

         string path_;
     };

     struct linked_node : public node_info
     {
                     linked_node();
                     linked_node(const linked_node&);
         linked_node& operator=(const linked_node&);

         string linkpath;
         bool queued;
         bool hidden;
     };

And then add the appropriate member functions - as needed.

Greg

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