有基类Customer如下,
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void logCall(const std::string& funcName);
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class Customer {
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public:
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//...
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Customer(const Customer& rhs);
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Customer& operator=(const Customer& rhs);
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//...
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private:
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std::string name;
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};
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Customer::Customer(const Customer & rhs)
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: name(rhs.name)
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{
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logCall("Customer copy constructor");
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}
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Customer& Customer::operator=(const Customer& rhs)
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{
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logCall("Customer copy assignment operator");
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name = rhs.name;
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return *this;
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}
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倘若该类被PrioCustomer类继承,其构造如下,
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class PrioCustomer: public Customer {
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public:
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//...
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PrioCustomer(const PrioCustomer& rhs);
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PrioCustomer& operator=(const PrioCustomer& rhs);
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//...
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private:
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int priority;
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};
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PrioCustomer::PrioCustomer(const PrioCustomer & rhs)
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: priority(rhs.priority);
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{
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logCall("PrioCustomer copy constructor");
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}
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PrioCustomer& PrioCustomer::operator=(const PrioCustomer & rhs)
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{
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logCall("PrioCustomer copy assignment operator");
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priority = rhs.priority;
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return *this;
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}
初看好像该copy构造函数没问题,但这里需要指明的是PrioCustomer类只复制了其自身定义的成员变量,并没有copy基类Customer的成员变量,
其基类Customer的成员变量只是以默认参数初始化之。
正确的构造应该,
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PrioCustomer::PrioCustomer(const PrioCustomer & rhs)
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: Customer(rhs),
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priority(rhs.priority);
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{
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logCall("PrioCustomer copy constructor");
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}
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PrioCustomer& PrioCustomer::operator=(const PrioCustomer & rhs)
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{
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logCall("PrioCustomer copy assignment operator");
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Customer::operator=(rhs);
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priority = rhs.priority;
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return *this;
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}
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