212 lines
4.6 KiB
Markdown
212 lines
4.6 KiB
Markdown
# 初始化列表与赋值
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- const成员的初始化只能在构造函数初始化列表中进行
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- 引用成员的初始化也只能在构造函数初始化列表中进行
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- 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
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## 类之间嵌套
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**第一种： 使用初始化列表。**
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```cpp
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class Animal {
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public:
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Animal() {
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std::cout << "Animal() is called" << std::endl;
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}
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Animal(const Animal &) {
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std::cout << "Animal (const Animal &) is called" << std::endl;
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}
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Animal &operator=(const Animal &) {
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std::cout << "Animal & operator=(const Animal &) is called" << std::endl;
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return *this;
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}
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~Animal() {
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std::cout << "~Animal() is called" << std::endl;
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}
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};
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class Dog {
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public:
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Dog(const Animal &animal) : __animal(animal) {
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std::cout << "Dog(const Animal &animal) is called" << std::endl;
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}
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~Dog() {
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std::cout << "~Dog() is called" << std::endl;
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}
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private:
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Animal __animal;
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};
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int main() {
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Animal animal;
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std::cout << std::endl;
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Dog d(animal);
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std::cout << std::endl;
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return 0;
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}
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```
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运行结果：
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```cpp
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Animal() is called
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Animal (const Animal &) is called
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Dog(const Animal &animal) is called
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~Dog() is called
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~Animal() is called
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~Animal() is called
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```
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依次分析从上到下：
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main函数中`Animal animal;`调用默认构造。
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`Dog d(animal);`等价于：
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```
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Animal __animal = animal;
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```
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实际上就是调用了拷贝构造，因此输出了：
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```
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Animal (const Animal &) is called
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```
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再然后打印Dog的构造函数里面的输出。
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最后调用析构，程序结束。
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**第二种：构造函数赋值来初始化对象。**
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构造函数修改如下：
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```cpp
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Dog(const Animal &animal) {
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__animal = animal;
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std::cout << "Dog(const Animal &animal) is called" << std::endl;
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}
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```
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此时输出结果：
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```
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Animal() is called
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Animal() is called
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Animal & operator=(const Animal &) is called
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Dog(const Animal &animal) is called
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~Dog() is called
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~Animal() is called
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~Animal() is called
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```
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于是得出：
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当调用`Dog d(animal);`时，等价于：
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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。
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```cpp
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// 假设之前已经有了animal对象
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Animal __animal;
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__animal = animal;
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```
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> 小结
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通过上述我们得出如下结论：
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- **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**
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- **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。**
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## 无默认构造函数的继承关系中
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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：
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```cpp
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class Animal {
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public:
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Animal(int age) {
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std::cout << "Animal(int age) is called" << std::endl;
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}
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Animal(const Animal & animal) {
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std::cout << "Animal (const Animal &) is called" << std::endl;
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}
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Animal &operator=(const Animal & amimal) {
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std::cout << "Animal & operator=(const Animal &) is called" << std::endl;
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return *this;
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}
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~Animal() {
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std::cout << "~Animal() is called" << std::endl;
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}
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};
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class Dog : Animal {
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public:
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Dog(int age) : Animal(age) {
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std::cout << "Dog(int age) is called" << std::endl;
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}
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~Dog() {
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std::cout << "~Dog() is called" << std::endl;
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}
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};
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```
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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？
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去掉初始化列表
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```
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Dog(int age) {
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std::cout << "Dog(int age) is called" << std::endl;
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}
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```
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运行程序：
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```
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error: no matching function for call to ‘Animal::Animal()’
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```
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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。
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## 类中const数据成员、引用数据成员
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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！
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例如：在上述的Animal中添加私有成员，并修改构造函数：
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```cpp
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class Animal {
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public:
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Animal(int age,std::string name) {
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std::cout << "Animal(int age) is called" << std::endl;
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}
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private:
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int &age_;
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const std::string name_;
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};
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```
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报下面错误：
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```cpp
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error: uninitialized reference member in ‘int&’
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```
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应该改为下面：
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```cpp
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Animal(int age, std::string name) : age_(age), name_(name) {
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std::cout << "Animal(int age) is called" << std::endl;
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}
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```
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