ValExi's questions

从上一个问题中我知道，在~base()调用派生类时，它首先构造一个临时基对象，然后operator~()在临时对象上调用临时基，最后销毁临时对象。

但是当基类没有时operator~，在派生类中调用base::~base()仍然可以调用基类的析构函数。

为什么当基类有时会发生错误operator~？

#include <iostream>
using namespace std;
class base
{
public:
    base(){cout << "constructor base \n";}
    ~base(){cout << "destructor base \n";}
    void operator~(){cout << "operator ~ \n";}
};

class derived : public base
{
public:
    derived(){cout << "constructor derived \n";}
    ~derived(){base::~base();cout << "destructor derived \n";}
};

编译器版本信息：

g++.exe (x86_64-posix-seh-rev0, Built by MinGW-W64 project) 8.1.0
Copyright (C) 2018 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

错误：

cmd /c chcp 65001>nul && C:\mingw64\bin\g++.exe -fdiagnostics-color=always -g "D:\Valkyrie-text\simple text\text.cpp" -o "D:\Valkyrie-text\simple text\text.exe"
D:\Valkyrie-text\simple text\text.cpp: In destructor 'derived::~derived()':
D:\Valkyrie-text\simple text\text.cpp:15:28: error: no matching function for call to 'derived::~derived()'
     ~derived(){base::~base();cout << "destructor derived \n";}
                            ^
D:\Valkyrie-text\simple text\text.cpp:7:5: note: candidate: 'base::~base()'
     ~base(){cout << "destructor base \n";}
     ^
D:\Valkyrie-text\simple text\text.cpp:7:5: note:   candidate expects 1 argument, 0 provided

在 C++ 中，为什么派生类不能不使用范围解析运算符而直接显式地调用其基类的析构函数？

我听说在C++中，所有的析构函数都会被编译器处理成“destructor”，也就是说基类和派生类的析构函数同名，从而导致重定义（隐藏）。

这是正确的吗？

#include <iostream>
 
struct base {
    base() { std::cout << "base default \n"; }
    ~base() { std::cout << "~base \n"; }
};
 
struct derived : base {
    derived() { std::cout << "derived default \n"; }
    ~derived()
    {
        //~base();No
        base::~base();
        std::cout << "~derived \n";
    }
};

我知道在实践中几乎没有必要这样做，我只是想了解原理......