技術標籤：C++學習c++

#include <iostream>

class test{
public:
    test(int a,int b):x(a),y(b){
        std::cout<<"test t:"<<x<<"  "<<y<<std::endl;
    }

    ~test() {
        std::cout << "~test()" << std::endl;        
    }
 

public:
    int x;
    int y;
};
//指標來定義形參
void alterTest0(test *p)
{
    p->x=100;
    p->y=100;
}
//引用來定義形參
void alterTest1(test &p)
{
    p.x=200;
    p.y=200;
}
//物件本身來定義形參
void alterTest2(test p)
{
    p.x=300;
    p.y=300;
}

int main(int argc, char* argv[])
{  
    test t(10,10);
    std::cout<<"test t:"
 
<<t.x<<"  "<<t.y<<std::endl;
    std::cout<<"\n-----------------------------------------------------"<<std::endl;
    alterTest2(t);
    std::cout<<"test t 物件本身做實參:"<<t.x<<"  "<<t.y<<std::endl;

    std:
 
:cout<<"\n-----------------------------------------------------"<<std::endl;
    alterTest0(&t);
    std::cout<<"test t 物件指標做實參:"<<t.x<<"  "<<t.y<<std::endl;

    std::cout<<"\n-----------------------------------------------------"<<std::endl;
    alterTest1(t);
    std::cout<<"test t 物件引用做實參:"<<t.x<<"  "<<t.y<<std::endl;
    
   return 0;
}

輸出：
說明問題：

1. 物件傳參：不會改變物件成員的值，（為什麼呼叫一次解構函式？）
2. 物件指標和物件引用都會改變物件成員的值