初始化std::array类型的类继承成员var的最佳方法?

有两个选项,但一个依赖于运行时大小验证。注意,在我的示例中,后者相当于一个cast。铸造有什么问题?

#include <cassert>
#include <algorithm>
#include <array>
#include <initializer_list>
#include <iostream>

struct A {
  typedef std::array<char const*, 3> T;
  T data_;

  A(std::initializer_list<char const*> const& data) {
    assert(data.size() <= data_.size());  // or ==
    // of course, use different error handling as appropriate
    std::copy(data.begin(), data.end(), data_.begin());
    std::fill(data_.begin() + data.size(), data_.end(), nullptr);
  }

  A(T const& data) : data_ (data) {}
};

int main() {
  A a ({"a", "b"});
  std::cout << (void const*)a.data_[2] << '\n';

  A b ((A::T{"a", "b"}));  // might just be the compiler I tested, but the
                           // extra parens were required; could be related
                           // to "the most vexing parse" problem
  std::cout << (void const*)b.data_[2] << '\n';

  return 0;
}

但是,对于每个Student对象,这些数据看起来是相同的。为什么不使用虚拟方法或将共享对象传递给基对象呢?你可以复制那个物体, 下面是与当前代码等效的代码,或者要求它比当前代码长,并存储指针/引用。

struct Student : Entity<int, string, string, bool> {
  Student() : Entity<int, string, string, bool>("student", entity_data_) {}
  // the template arguments are still required in 0x, I believe
private:
  static Entity<int, string, string, bool>::source_names_t entity_data_;
}