复制构造函数的const对象

Question

我正在阅读有关复制构造函数的内容。 任何机构可以告诉我什么是在下面的语句中发生

class Base { 
public: 
Base() {cout << "Base constructor";} 
Base(const Base& a) {cout << "copy constructor with const arg";} 
Base(Base& a) {cout << "copy constructor with non-const arg"; return a;} 
const Base& operator=(Base &a) {cout << "assignment operator with non-const arg"; return a;} 
} 

void main() 
{ 
    Base a; 
    Base b = Base(); // This is neither calling copy constructor nor assignment operator. 
} 

请告诉我什么是在“基地B =基地（）”语句发生。

Answer 1

拷贝构造函数将在三个casess被称为：

When an object is returned by value 
When an object is passed (to a function) by value as an argument 
When an object is thrown 
When an object is caught 
When an object is placed in a brace-enclosed initializer list 

分配opertator将被调用时，下面：

B b; 
b=a; 

所以你的语句：

Base b = Base(); 

不适合任何上述的。