初始化列表与赋值

  • const成员的初始化只能在构造函数初始化列表中进行
  • 引用成员的初始化也只能在构造函数初始化列表中进行
  • 对象成员(对象成员所对应的类没有默认构造函数)的初始化,也只能在构造函数初始化列表中进行

类之间嵌套

第一种: 使用初始化列表。

class Animal {
public:
    Animal() {
        std::cout << "Animal() is called" << std::endl;
    }

    Animal(const Animal &) {
        std::cout << "Animal (const Animal &) is called" << std::endl;
    }

    Animal &operator=(const Animal &) {
        std::cout << "Animal & operator=(const Animal &) is called" << std::endl;
        return *this;
    }

    ~Animal() {
        std::cout << "~Animal() is called" << std::endl;
    }
};

class Dog {
public:
    Dog(const Animal &animal) : __animal(animal) {
        std::cout << "Dog(const Animal &animal) is called" << std::endl;
    }

    ~Dog() {
        std::cout << "~Dog() is called" << std::endl;
    }

private:
    Animal __animal;
};

int main() {
    Animal animal;
    std::cout << std::endl;
    Dog d(animal);
    std::cout << std::endl;
    return 0;
}

运行结果:

Animal() is called

Animal (const Animal &) is called
Dog(const Animal &animal) is called

~Dog() is called
~Animal() is called
~Animal() is called

依次分析从上到下:

main函数中Animal animal;调用默认构造。

Dog d(animal);等价于:

Animal __animal = animal;

实际上就是调用了拷贝构造,因此输出了:

Animal (const Animal &) is called

再然后打印Dog的构造函数里面的输出。

最后调用析构,程序结束。

第二种:构造函数赋值来初始化对象。

构造函数修改如下:

Dog(const Animal &animal) {
    __animal = animal;
    std::cout << "Dog(const Animal &animal) is called" << std::endl;
}

此时输出结果:

Animal() is called

Animal() is called
Animal & operator=(const Animal &) is called
Dog(const Animal &animal) is called

~Dog() is called
~Animal() is called
~Animal() is called

于是得出:

当调用Dog d(animal);时,等价于:

先定义对象,再进行赋值,因此先调用了默认构造,再调用=操作符重载函数。

// 假设之前已经有了animal对象
Animal __animal;
__animal = animal;

小结

通过上述我们得出如下结论:

  • 类中包含其他自定义的class或者struct,采用初始化列表,实际上就是创建对象同时并初始化
  • 而采用类中赋值方式,等价于先定义对象,再进行赋值,一般会先调用默认构造,在调用=操作符重载函数。

无默认构造函数的继承关系中

现考虑把上述的关系改为继承,并修改Animal与Dog的构造函数,如下代码:

class Animal {
public:
    Animal(int age) {
        std::cout << "Animal(int age) is called" << std::endl;
    }

    Animal(const Animal & animal) {
        std::cout << "Animal (const Animal &) is called" << std::endl;
    }

    Animal &operator=(const Animal & amimal) {
        std::cout << "Animal & operator=(const Animal &) is called" << std::endl;
        return *this;
    }

    ~Animal() {
        std::cout << "~Animal() is called" << std::endl;
    }
};

class Dog : Animal {
public:
    Dog(int age) : Animal(age) {
        std::cout << "Dog(int age) is called" << std::endl;
    }

    ~Dog() {
        std::cout << "~Dog() is called" << std::endl;
    }

};

上述是通过初始化列表给基类带参构造传递参数,如果不通过初始化列表传递,会发生什么影响?

去掉初始化列表

Dog(int age)  {
    std::cout << "Dog(int age) is called" << std::endl;
}

运行程序:

error: no matching function for call to ‘Animal::Animal()’

由于在Animal中没有默认构造函数,所以报错,遇到这种问题属于灾难性的,我们应该尽量避免,可以通过初始化列表给基类的构造初始化。

类中const数据成员、引用数据成员

特别是引用数据成员,必须用初始化列表初始化,而不能通过赋值初始化!

例如:在上述的Animal中添加私有成员,并修改构造函数:

class Animal {
public:
    Animal(int age,std::string name) {
        std::cout << "Animal(int age) is called" << std::endl;
    }
private:
    int &age_;
    const std::string name_;
};

报下面错误:

error: uninitialized reference member in int&

应该改为下面:

Animal(int age, std::string name) : age_(age), name_(name) {
    std::cout << "Animal(int age) is called" << std::endl;
}