具有虚拟赋值运算符的多态性

Polymorphism with virtual assignment operator

本文关键字:多态性 赋值运算符 虚拟      更新时间:2023-10-16

我不明白为什么下面的代码调用基的类 (=A) 实现

A::operator=(A&)

(由输出行 15. 和 17 指示。) 每当作为参数传入的引用属于派生类类型时。由于所有指针都指向派生类的对象,因此我希望所有调用都具有相同的行为(=从派生类 B 调用运算符)。

执行下面列出的代码时输出:

  1. A::Ctor objID=1280 text=BObject1
  2. B::Ctor objID=1280 text=BObject1
  3. A::Ctor objID=3279 text=BObject2
  4. B::Ctor objID=3279 text=BObject2
  5. --- ptrA->print():
  6. B::p rint() objID=1280 text=BObject1
  7. B::p rint() objID=3279 text=BObject2
  8. --- *ptrAObj1 = *ptrAObj2:
  9. B::运算符=(A&) objID=1280 text=BObject1
  10. --- ptrAObj1->operator=(*ptrAObj2):
  11. B::运算符=(A&) objID=1280 text=BObject2
  12. --- *ptrBObj1 = *ptrAObj2:
  13. B::运算符=(A&) objID=1280 text=BObject2
  14. --- *ptrBObj1 = *ptrBObj2:
  15. A:
  16. :运算符=(A&) objID=1280 text=BObject2
  17. --- ptrBObj1->operator=(*ptrBObj2):
  18. A:
  19. :运算符=(A&) objID=1280 text=BObject2
  20. --- ptrAObj1->operator=(*ptrBObj2):
  21. B::运算符=(A&) objID=1280 text=BObject2
  22. ---- 拆解()
  23. B::D tor objID=1280 text=BObject2
  24. A::D tor objID=1280 text=BObject2
  25. B::D tor objID=3279 text=BObject2
  26. A::D tor objID=3279 text=BObject2

法典:

#include "gtest/gtest.h"
#include <stdio.h>
#include <string>
#include <ctime>
class A
{
public:
A(const std::string& sLabel=""):
m_sText(sLabel)
{
std::srand(std::time(nullptr)); // use current time as seed for random generator
int random_no = (std::rand()+(m_iObjectCounter*1999)) % 9999;
m_iObjectID = random_no;
m_iObjectCounter++;
std::cout << "A::Ctor objID=" << m_iObjectID << " text=" << m_sText << std::endl;
}
virtual ~A()
{
std::cout << "A::Dtor objID=" << m_iObjectID << " text=" << m_sText << std::endl;
}
virtual void print() const
{
std::cout << "A::print() objID=" << m_iObjectID << " text=" << m_sText << std::endl;
}
virtual int operator=(A& rIn)
{
std::cout << "A::operator=(A&) objID=" << m_iObjectID << " text=" << m_sText << std::endl;
m_sText = rIn.m_sText;
return EXIT_SUCCESS;
}
std::string m_sText;
int m_iObjectID;
static int m_iObjectCounter;
};
int A::m_iObjectCounter=0;

class B : public A
{
public:
B(const std::string& sLabel = "") :
A(sLabel)
{
std::cout << "B::Ctor objID=" << m_iObjectID << " text=" << m_sText << std::endl;
}
virtual ~B()
{
std::cout << "B::Dtor objID=" << m_iObjectID << " text=" << m_sText << std::endl;
}
virtual void print() const
{
std::cout << "B::print() objID=" << m_iObjectID << " text=" << m_sText << std::endl;
}
virtual int operator=(A& rIn)
{
std::cout << "B::operator=(A&) objID=" << m_iObjectID << " text=" << m_sText << std::endl;
m_sText = rIn.m_sText;
return EXIT_SUCCESS;
}
};

class TestAssignmentOperator : public ::testing::Test {
public:
TestAssignmentOperator() :
ptrAtoBObject1(NULL),
ptrAtoBObject2(NULL)
{
ptrAtoBObject1 = new B("BObject1");
ptrAtoBObject2 = new B("BObject2");
}
void SetUp(){}
void TearDown() {
std::cout << " ---- TearDown()" << std::endl;
delete ptrAtoBObject1;
delete ptrAtoBObject2;
}
~TestAssignmentOperator(){}
A* ptrAtoBObject1;
A* ptrAtoBObject2;
};

TEST_F(TestAssignmentOperator, Test1)
{
std::cout << " --- ptrA->print():" << std::endl;
ptrAtoBObject1->print();
ptrAtoBObject2->print();

int iError = EXIT_FAILURE;
std::cout << " --- *ptrAObj1 = *ptrAObj2:" << std::endl;
iError = *ptrAtoBObject1 = *ptrAtoBObject2;

std::cout << " --- ptrAObj1->operator=(*ptrAObj2):" << std::endl;
iError = ptrAtoBObject1->operator=(*ptrAtoBObject2);

std::cout << " --- *ptrBObj1 = *ptrAObj2:" << std::endl;
B* ptrBToObject1 = dynamic_cast<B*>(ptrAtoBObject1);
iError = *ptrBToObject1 = *ptrAtoBObject2;

std::cout << " --- *ptrBObj1 = *ptrBObj2:" << std::endl;
B* ptrBtoBObject2 = dynamic_cast<B*>(ptrAtoBObject2);
*ptrBToObject1 = *ptrBtoBObject2;

std::cout << " --- ptrBObj1->operator=(*ptrBObj2):" << std::endl;
ptrBToObject1->operator=(*ptrBtoBObject2);

std::cout << " --- ptrAObj1->operator=(*ptrBObj2):" << std::endl;
ptrAtoBObject1->operator=(*ptrBtoBObject2);
}

int main(int argc, char **argv)
{
::testing::InitGoogleTest(&argc, argv);
return RUN_ALL_TESTS();
}

我相信这是因为 B 类中的virtual int operator=(A& rIn)不是真正的默认operator=,因此编译器正在为您创建一个B& operator=(const B& other)并使用该。

除此之外,覆盖虚拟运算符=是不够的,还必须定义常规运算符=。

当然,最好使用 C++11 功能(如override关键字)来保证您确实编写了您想要的函数签名,而不是创建新的虚拟函数。

在这两种情况下,您都调用默认B::operator=(const B&)。然后,其默认实现调用A::operator=(A&)