您可以为具有 const 成员的联合编写一个复制构造函数吗？

Question

Suppose I have a struct that contains a union with const members, like so:

struct S
{
  // Members

  const enum { NUM, STR } type;

  union
  {
    const int a;
    const std::string s;
  };

  // Constructors

  S(int t_a) : type(NUM), a(t_a);

  S(const std::string & t_s) : type(STR), s(t_s);

};

So far, so good. But now say I want to write a copy-constructor for this type.

It doesn't seem like this involves doing anything nefarious, but since I need to initialize the const members in member initializers I don't see how to do this based on logic that depends on the type member.

Questions:

• Is it possible to write this constructor?

• If not, is this essentially a syntactic oversight, or is there some fundamental reason that the language can't support such a thing?

Answer 1

Yes, it is possible to write copy constructor here. Actually it is already done inside std::variant implementation, which shall support const -types among others. So your class S can be replaced with

using S = std::variant<const int, const std::string>;

But if for dome reason you cannot use std::variant then copy-constructor can be written using std::construct_at function as follows:

#include <string>

struct S {
  const enum { NUM, STR } type;

  union {
    const int a;
    const std::string s;
  };

  S(int t_a) : type(NUM), a(t_a) {}
  S(const std::string & t_s) : type(STR), s(t_s) {}
  S(const S & rhs) : type(rhs.type) {
      if ( type == NUM ) std::construct_at( &a, rhs.a );
      if ( type == STR ) std::construct_at( &s, rhs.s );
  }
  ~S() {
      if ( type == STR ) s.~basic_string();
  }
};

int main() {
    S s(1);
    S u = s;

    S v("abc");
    S w = v;
}

Demo: https://gcc.godbolt.org/z/TPe8onhWs