c++ - 复制构造函数 C++

Question

抱歉，我正在努力掌握复制构造函数，我想知道为什么仅当我按值“op.return_Value(op)< 调用函数中的对象时才调用复制构造函数

class operation{
   public:
       int add(int x,int y){
           *total=x+y;
       return (*total);
       }

       operation(){
       cout<<"This is the constructor"<<endl;
       total=new int;
       }
       operation(const operation &op){
       cout<<"This is the copy of the start"<<endl;
       total=new int;
       *total=*op.total;
       }

       ~operation(){
       cout<<"This is the end";
       delete total;
       }

        int *total;

       int return_Value(operation op){
        return *total;
       }

};

class child_operation:operation{

   public:

       int sub(int x,int y){
           *total=x-y;
       return(*total);
       }

};



int main()
{
   operation op;
   child_operation op1;

   cout<<op.add(5,6)<<endl<<op1.sub(6,5)<<endl;
   cout<<op.return_Value(op)<<endl;
}

复制构造函数基本上以哪些方式被调用？

Answer 1

它非常简单，并在标准 (12.8.1) 中明确说明:

A class object can be copied in two ways, by initialization (including for function argument passing and for function value return) and by assignment. Conceptually, these two operations are implemented by a copy constructor and copy assignment operator.

因此，如果您要显式或隐式地初始化一个新对象(例如，通过复制将一个对象传递给函数，这是您的情况，或者从函数返回一个对象)，复制构造函数被调用:

MyClass a = b;//initialization, copy constructor is called
//or
void foo(MyClass a){}//a is passed by value, a copy will be made, and copy constructor will be called
//or
MyClass foo()
{
    MyClass result;
    return result;//a copy of result will be returned, and copy constructor will be called
}

相反，如果您要进行赋值，则会调用 operator=，例如

MyClass x;
MyClass y;
y = x;//assignmanet