c++ - 原始类型的强 typedef (BOOST_STRONG_TYPEDEF 不会削减它)

Question

我之前使用过BOOST_STRONG_TYPEDEF，主要是使用std::string，得到了满意的结果:

#include <boost/serialization/strong_typedef.hpp>
#include <iostream>

BOOST_STRONG_TYPEDEF(std::string, TIMER_ID)
BOOST_STRONG_TYPEDEF(std::string, PROCESS_ID)

int main()
{
    TIMER_ID t_id("Timer");
    PROCESS_ID p_id("Process");

    if (t_id == p_id)
        std::cout << "They are equal!" << std::endl;
}

前面的代码无法按预期编译:

In file included from /usr/include/boost/serialization/strong_typedef.hpp:26:0,
                 from types.cpp:1:
/usr/include/boost/operators.hpp: In instantiation of ‘bool boost::operator==(const std::__cxx11::basic_string<char>&, const PROCESS_ID&)’:
types.cpp:12:14:   required from here
/usr/include/boost/operators.hpp:144:64: error: no match for ‘operator==’ (operand types are ‘const PROCESS_ID’ and ‘const std::__cxx11::basic_string<char>’)
      friend bool operator==(const U& y, const T& x) { return x == y; }

但是，这段代码编译得很好:

#include <boost/serialization/strong_typedef.hpp>
#include <iostream>

BOOST_STRONG_TYPEDEF(unsigned int, TIMER_ID)
BOOST_STRONG_TYPEDEF(unsigned int, PROCESS_ID)

int main()
{
    TIMER_ID t_id(12);
    PROCESS_ID p_id(12);

    if (t_id == p_id)
    {
        std::cout << "They are equal!" << std::endl;
        std::cout << "Their sum is " << t_id + p_id << std::endl;
    }
}

这看起来一点也不强啊!我希望在没有 static_cast 的情况下无法比较或添加两种不同类型的对象。

• 为什么会发生这种情况？
• 如何在不为每种类型手动创建类的情况下实现基本类型的类型安全？

Answer 1

• 正在阅读 http://www.boost.org/doc/libs/1_63_0/libs/serialization/doc/strong_typedef.html，看来您使用 boost 的示例可以编译，因为 boost 创建的类型可以替代具有可比性的原始类型。

• 来自 fluentcpp 的 Johnathan Boccara 在其 github 上提供了强类型的实现，这应该是您想要的:

#include <cassert>
#include "NamedType/named_type.hpp"

int main() {
  using TIMER_ID =
      fluent::NamedType<unsigned int, struct TimerIdTag, fluent::Comparable>;
  using PROCESS_ID =
      fluent::NamedType<unsigned int, struct ProcessIdTag, fluent::Comparable>;

  TIMER_ID a(123);
  PROCESS_ID b(456);

  assert(a == a);
  // assert(a == b); doesn't compile
  return 0;
}