我试图向每个添加一个副本构造函数以运行主要功能。我现在实现的内容现在会打印出“ b2.valuea = -858993460 b2.valueb = 10”,因此它正确读取了valueb,但显然valuea出了问题。忠告?
#include "stdafx.h"
#include <iostream>
using namespace std;
class A
{
int valuea;
public:
int getValuea() const { return valuea; }
void setValuea(int x) { valuea = x; }
// copy constructor
A() {}
A(const A& original)
{
valuea = original.valuea;
}
};
class B : public A
{
int valueb;
public:
int getValueb() const { return valueb; }
void setValueb(int x) { valueb = x; }
// copy constructor
B(){}
B(const B& original)
{
valueb = original.valueb;
}
};
int main()
{
B b1;
b1.setValuea(5);
b1.setValueb(10);
B b2(b1);
cout << "b2.valuea = " << b2.getValuea() << "b2.valueb = " <<
b2.getValueb() << endl;
}
最佳答案
您没有在派生的副本构造函数中调用基类的副本构造函数。
像这样更改它:
B(const B& original) : A(original){
valueb = original.valueb;
}
它会输出
b2.valuea = 5
b2.valueb = 10