C ++中默认副本构造函数的怪异行为

Question

I've encountered a weird behavior during learning about implicit copy constructors in C++. This is about this bunch of code:

#include <iostream>

class Person
{
public:
    char* name;
    int age;

    Person( const char* the_name, int the_age )
    {
        name = new char[strlen( the_name ) + 1];
        strcpy( name, the_name );
        age = the_age;
    }

    ~Person()
    {
        delete[] name;
    }
};

int main(){
    Person p1( "Tom", 22 );

    Person p2( p1 );
    p2.name = "Bill";

    // the output is: "Tom"
    std::cout << p1.name << std::endl;

    std::cin.get();
}

By default, copying an object means copying its members, therefore while creating p2 object using default copy constructor it should copy merely a pointer name , not the character array it points to. Therefore changing name member of object p2 should change the name member of object p1 . Thereupon p1.name should be "Bill" not "Tom" .

What's wrong in my reasoning? Why it's printing "Tom" ?

Answer 1

By default, copying an object means copying its members

Yes.

therefore while creating p2 object using default copy constructor it should copy merely a pointer name , not the character array it points to.

Correct.

Therefore changing name member of object p2 should change the name member of object p1 .

Nope. Yes, p2.name and p1.name point to the same memory location, but that doesn't mean that changing the pointer value of p1.name will change the value of p2.name . If that were int s, would you be surprised if changing p1.name has no effect p2.name ? Because it's the same thing here. You have two different variables, and changing one's value doesn't change the value of the other.

int a = 0;
int b = 1;

int* ptrA = &a;
int *ptrA2 = &a;

*ptrA = 1; // now a == 1
*ptrA2 = 4; // now a == 4

ptrA2 = &b;

*ptrA2 = 10; // now b == 10
*ptrA = 3; // now a == 3, a != 10