c – 如何在遵守得墨忒耳法的同时延迟加载？

template <typename T>
class Lazy {
public:
  explicit Lazy(T const& t): _loader(),_item(t) {}
  explicit Lazy(T&& t): _loader() _item(t) {}
  explicit Lazy(std::function<T()> l): _loader(l),_item() {}

  T& operator*() { return this->get(); }
  T const& operator*() const { return this->get(); }

  T* operator->() { return &this->get(); }
  T const* operator->() const { return &this->get(); }

private:
  T& get() { if (not _item) { _item = _loader(); } return *_item; }
  T const& get() const { if (not _item) { _item = _loader(); } return *_item; }

  std::function<T()> _loader;
  mutable boost::optional<T> _item;
}; // class Lazy

只要完成,该类就可以保证加载器执行一次.如果未完成,则将在下次访问时重试执行.它不是线程安全的.

用法：

// Output prime numbers as text
bool isPrime(int);
std::string toEnglish(int);

void print(int i,Lazy<std::string> const& heavy) {
    if (not isPrime(i)) { return; }

    std::cout << *heavy << "n";
}

int main(int argc,char* argv[]) {
    if (argc < 2) { return 1; }

    int max = atoi(argv[1]);
    for (int i = 2; i <= max; ++i) {
        print(i,[=](){ return toEnglish(i); });
              // ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++11 lambda syntax
    }
}

在这里,toEnglish只用于素数.