使用BOOST_CLASS_EXPORT导出模板类，如何无需类型别名实现？

问题描述 #

直接编写BOOST_CLASS_EXPORT(DerivedA<double, float>)时会触发编译错误，编译器认为宏BOOST_CLASS_EXPORT被传入2个参数，但该宏实际仅接受1个参数，且无法将其识别为类型。而通过定义类型别名using alias = DerivedA<double, float>; BOOST_CLASS_EXPORT(alias);可正常编译运行，现寻求无需类型别名的更简洁解决方案。

错误信息 #

error: macro "BOOST_CLASS_EXPORT" passed 2 arguments, but takes just 1
   35 | BOOST_CLASS_EXPORT(DerivedA<double, float>) // ERROR
      |                                           ^
In file included from /opt/boost-1.79.0-r7de4sm2snkkg5mu5iamq2vdfpwrloi4/include/boost/serialization/export.hpp:215: note: macro "BOOST_CLASS_EXPORT" defined here
  215 | #define BOOST_CLASS_EXPORT(T)                   \
      | 
/home/test/main.cc:35:1: error: ‘BOOST_CLASS_EXPORT’ does not name a type
   35 | BOOST_CLASS_EXPORT(DerivedA<double, float>) // ERROR
      | ^~~~~~~~~~~~~~~~~~

代码示例 #

#include <iostream>
#include <memory>
#include <boost/archive/text_oarchive.hpp>
#include <boost/archive/text_iarchive.hpp>
#include <boost/serialization/shared_ptr.hpp>
#include <boost/serialization/export.hpp>

// Base class
template<typename T, typename U>
class Base {
public:
    virtual ~Base() = default;

    template<class Archive>
    void serialize(Archive& ar, const unsigned int version) {
        // No members to serialize
    }
};

template<typename T, typename U>
class DerivedA : public Base<T, U> {
public:
    T valueA;
    U valueB;

    DerivedA() = default;

    template<class Archive>
    void serialize(Archive& ar, const unsigned int version) {
        ar & boost::serialization::base_object<Base<T, U>>(*this);
        ar & valueA;
        ar & valueB;
    }
};
BOOST_CLASS_EXPORT(DerivedA<double, float>) // ERROR
// using alias = DerivedA<double, float>;
// BOOST_CLASS_EXPORT(alias); // WORKS


// Struct holding a shared_ptr to Base
template<typename T, typename U>
struct MyStruct {
    std::shared_ptr<Base<T, U>> ptr;

    template<class Archive>
    void serialize(Archive& ar, const unsigned int version) {
        ar &ptr; 
    }
};

int main() {
    MyStruct<double, float> original;
    original.ptr = std::make_shared<DerivedA<double, float>>(); // Store DerivedA
    return 0;
}

解决方案 #

核心问题是C++预处理器解析宏参数时，会把尖括号内的逗号视为宏的参数分隔符，导致DerivedA<double, float>被拆成两个参数传入宏。无需类型别名的简洁解决办法是用双层括号包裹模板实例化类型：

BOOST_CLASS_EXPORT((DerivedA<double, float>))

原理说明 #

双层括号会让预处理器将内部的DerivedA<double, float>当作一个完整的参数整体解析，尖括号内的逗号不会被误判为宏的参数分隔符，从而匹配宏BOOST_CLASS_EXPORT(T)仅接受1个参数的定义，编译即可正常通过。

内容的提问来源于stack exchange，提问作者Mathieu