基类含unique_ptr时，引用类型dynamic_cast为何编译失败？

问题场景 #

基类包含接收基类引用的纯虚函数（用于拷贝逻辑），派生类中尝试将该引用转为派生类类型，原本正常；但在基类添加Base类型的unique_ptr成员后，引用类型的dynamic_cast编译失败，改用指针类型则正常。注释掉unique_ptr成员或引用转换的代码时，程序可正常编译运行。

示例代码 #

#include <memory>
#include <iostream>

class Base {
public:
    Base() {}
    void hello() const { std::cout << "hello world" << std::endl; }
protected:
    virtual void cvt( Base& ) const = 0;
    std::unique_ptr<Base> any_member_unique_ptr;                    // "A"
};

class Derived : public Base {
public:
    Derived() { std::cout << "new Derived" << std::endl; }
    void cvt( Base& in ) const override {
        const auto cast_fail = dynamic_cast<const Derived&>( in );  // "B"
        const auto cast_okay = dynamic_cast<const Derived*>( &in );
        hello();
    }
};

int main(int argc, char *argv[])
{
    Derived D, P;
    D.cvt( P );
}

编译错误信息 #

tt.cpp: In member function ‘virtual void Derived::cvt(Base&) const’:
tt.cpp:17:69: error: use of deleted function ‘Derived::Derived(const Derived&)’
   17 |      const auto cast_fail = dynamic_cast<const Derived&>( in );  // "B"
      |                                                              ^

tt.cpp:13:7: note: ‘Derived::Derived(const Derived&)’ is implicitly deleted because the default definition would be ill-formed:
   13 | class Derived : public Base {
      |       ^~~~~~~
tt.cpp:13:7: error: use of deleted function ‘Base::Base(const Base&)’
tt.cpp:4:7: note: ‘Base::Base(const Base&)’ is implicitly deleted because the default definition would be ill-formed:
    4 | class Base {
      |       ^~~~
tt.cpp:4:7: error: use of deleted function ‘std::unique_ptr<_Tp, _Dp>::unique_ptr(const std::unique_ptr<_Tp, _Dp>&) [with _Tp = Base; _Dp = std::default_delete<Base>]’
In file included from /usr/include/c++/11/memory:76,
                 from tt.cpp:1:
/usr/include/c++/11/bits/unique_ptr.h:468:7: note: declared here
  468 |       unique_ptr(const unique_ptr&) = delete;
      |       ^~~~~~~~~~
make: *** [<builtin>: tt] Error 1

原因解析 #

1. 引用转换的隐式拷贝陷阱：dynamic_cast<const Derived&>(in)返回的是引用，但你写的const auto cast_fail = ...中，auto会推导为const Derived（值类型），这意味着代码试图用转换后的引用拷贝构造一个新的Derived对象。
2. 拷贝构造被删除的连锁反应：基类中的std::unique_ptr<Base>成员的拷贝构造函数是被显式删除的，因此基类的默认拷贝构造函数会被隐式删除；派生类继承了这一特性，其默认拷贝构造函数也会被隐式删除。当代码尝试拷贝构造Derived对象时，就会触发编译错误。
3. 指针转换无拷贝行为：指针版的dynamic_cast返回的是指针，const auto cast_okay推导为const Derived*，只是赋值指针地址，不需要拷贝构造对象，因此不会触发错误。

解决方案 #

将引用转换的结果声明为引用类型，避免触发拷贝构造：

// 方式1：显式声明引用类型
const Derived& cast_ok = dynamic_cast<const Derived&>(in);
// 方式2：用auto&推导引用类型
const auto& cast_ok = dynamic_cast<const Derived&>(in);

修改后，代码只是将引用绑定到原对象上，不会触发拷贝构造，即可正常编译运行。

内容的提问来源于stack exchange，提问作者rand'Chris