Unions as Base Class

Question

The standard defines that Unions cannot be used as Base class, but is there any specific reasoning for this? As far as I understand Unions can have constructors, destructors, also member variables, and methods to operate on those varibales. In short a Union can encapsulate a datatype and state which might be accessed through member functions. Thus it in most common terms qualifies for being a class and if it can act as a class then why is it restricted from acting as a base class?

Edit: Though the answers try to explain the reasoning I still do not understand how Union as a Derived class is worst than when Union as just a class. So in hope of getting more concrete answer and reasoning I will push this one for a bounty. No offence to the already posted answers, Thanks for those!

Answer 1

Tony Park gave an answer which is pretty close to the truth. The C++ committee basically didn't think it was worth the effort to make unions a strong part of C++, similarly to the treatment of arrays as legacy stuff we had to inherit from C but didn't really want.

Unions have problems: if we allow non-POD types in unions, how do they get constructed? It can certainly be done, but not necessarily safely, and any consideration would require committee resources. And the final result would be less than satisfactory, because what is really required in a sane language is discriminated unions, and bare C unions could never be elevated to discriminated unions in way compatible with C (that I can imagine, anyhow).

To elaborate on the technical issues: since you can wrap a POD-component only union in a struct without losing anything, there's no advantage allowing unions as bases. With POD-only union components, there's no problem with explicit constructors simply assigning one of the components, nor with using a bitblit (memcpy) for compiler generated copy constructor (or assignment).

Such unions, however, aren't useful enough to bother with except to retain them so existing C code can be considered valid C++. These POD-only unions are broken in C++ because they fail to retain a vital invariant they possess in C: any data type can be used as a component type.

To make unions useful, we must allow constructable types as members. This is significant because it is not acceptable to merely assign a component in a constructor body, either of the union itself, or any enclosing struct: you cannot, for example, assign a string to an uninitialised string component.

It follows one must invent some rules for initialising union component with mem-initialisers, for example:

union X { string a; string b; X(string q) : a(q) {} };

But now the question is: what is the rule? Normally the rule is you must initialise every member and base of a class, if you do not do so explicitly, the default constructor is used for the remainder, and if one type which is not explicitly initialised does not have a default constructor, it's an error [Exception: copy constructors, the default is the member copy constructor].

Clearly this rule can't work for unions: the rule has to be instead: if the union has at least one non-POD member, you must explicitly initialise exactly one member in a constructor. In this case, no default constructor, copy constructor, assignment operator, or destructor will be generated and if any of these members are actually used, they must be explicitly supplied.

So now the question becomes: how would you write, say, a copy constructor? It is, of course quite possible to do and get right if you design your union the way, say, X-Windows event unions are designed: with the discriminant tag in each component, but you will have to use placement operator new to do it, and you will have to break the rule I wrote above which appeared at first glance to be correct!

What about default constructor? If you don't have one of those, you can't declare an uninitialised variable.

There are other cases where you can determine the component externally and use placement new to manage a union externally, but that isn't a copy constructor. The fact is, if you have N components you'd need N constructors, and C++ has a broken idea that constructors use the class name, which leaves you rather short of names and forces you to use phantom types to allow overloading to choose the right constructor .. and you can't do that for the copy constructor since its signature is fixed.

Ok, so are there alternatives? Probably, yes, but they're not so easy to dream up, and harder to convince over 100 people that it's worthwhile to think about in a three day meeting crammed with other issues.

It is a pity the committee did not implement the rule above: unions are mandatory for aligning arbitrary data and external management of the components is not really that hard to do manually, and trivial and completely safe when the code is generated by a suitable algorithm, in other words, the rule is mandatory if you want to use C++ as a compiler target language and still generate readable, portable code. Such unions with constructable members have many uses but the most important one is to represent the stack frame of a function containing nested blocks: each block has local data in a struct, and each struct is a union component, there is no need for any constructors or such, the compiler will just use placement new. The union provides alignment and size, and cast free component access. [And there is no other conforming way to get the right alignment!]

Therefore the answer to your question is: you're asking the wrong question. There's no advantage to POD-only unions being bases, and they certainly can't be derived classes because then they wouldn't be PODs. To make them useful, some time is required to understand why one should follow the principle used everywhere else in C++: missing bits aren't an error unless you try to use them.

Answer 2

Union is a type that can be used as any one of its members depending on which member has been set - only that member can be later read.

When you derive from a type the derived type inherits the base type - the derived type can be used wherever the base type could be. If you could derive from a union the derived class could be used (not implicitly, but explicitly through naming the member) wherever any of the union members could be used, but among those members only one member could be legally accessed. The problem is the data on which member has been set is not stored in the union.

To avoid this subtle yet dangerous contradiction that in fact subverts a type system deriving from a union is not allowed.

Answer 3

Bjarne Stroustrup said 'there seems little reason for it' in The Annotated C++ Reference Manual.

Answer 4

The title asks why unions can't be a base class, but the question appears to be about unions as a derived class. So, which is it?

There's no technical reason why unions can't be a base class; it's just not allowed. A reasonable interpretation would be to think of the union as a struct whose members happen to potentially overlap in memory, and consider the derived class as a class that inherits from this (rather odd) struct. If you need that functionality, you can usually persuade most compilers to accept an anonymous union as a member of a struct. Here's an example, that's suitable for use as a base class. (And there's an anonymous struct in the union for good measure.)

struct V3 {
    union {
        struct {
            float x,y,z;
        };
        float f[3];
    };
};

The rationale for unions as a derived class is probably simpler: the result wouldn't be a union. Unions would have to be the union of all their members, and all of their bases. That's fair enough, and might open up some interesting template possibilities, but you'd have a number of limitations (all bases and members would have to be POD -- and would you be able to inherit twice, because a derived type is inherently non-POD?), this type of inheritance would be different from the other type the language sports (OK, not that this has stopped C++ before) and it's sort of redundant anyway -- the existing union functionality would do just as well.

Stroustrup says this in the D&E book:

As with void *, programmers should know that unions ... are inherently dangerous, should be avoided wherever possible, and should be handled with special care when actually needed.

(The elision doesn't change the meaning.)

So I imagine the decision is arbitrary, and he just saw no reason to change the union functionality (it works fine as-is with the C subset of C++), and so didn't design any integration with the new C++ features. And when the wind changed, it got stuck that way.

Answer 5

I think you got the answer yourself in your comments on EJP's answer.

I think unions are only included in C++ at all in order to be backwards compatible with C. I guess unions seemed like a good idea in 1970, on systems with tiny memory spaces. By the time C++ came along I imagine unions were already looking less useful.

Given that unions are pretty dangerous anyway, and not terribly useful, the vast new opportunities for creating bugs that inheriting from unions would create probably just didn't seem like a good idea :-)

Answer 6

Here's my guess for C++ 03.

As per $9.5/1, In C++ 03, Unions can not have virtual functions. The whole point of a meaningful derivation is to be able to override behaviors in the derived class. If a union cannot have virtual functions, that means that there is no point in deriving from a union.

Hence the rule.

Answer 7

You can inherit the data layout of a union using the anonymous union feature from C++11.

#include <cstddef>

template <size_t N,typename T>
struct VecData {
  union {
    struct {
      float x;
      float y;
      float z;
    };
    float a[N];
  };
};

template <size_t N, typename T>
  class Vec : public VecData<N,T> {
    //methods..
};

In general its almost always better not work with unions directly but enclose them within a struct or class. Then you can base your inheritance off the struct outer layer and use unions within if you need to.