how default assignment operator works with string pointer

Question

Should the code fail during runtime? Visual studio executes it without failure

#include<iostream>
#include <string>
using namespace std;

class Boy
{
    string* _name;
public:
    Boy(string n) :_name(new string(n)) {}
    string getName() const { return *_name; }
};
int main() {
    Boy t("Tom");
    string str = "Tom2";
    t = str;
    cout << t.getName();
        return 0;
}

Answer 1

Should the code fail during runtime?

No, it shouldn't.

Let's dissect, what the following line does:

t = str;

There's a temporary Boy instance implicitely constructed. Thus it's merely the same as

t = Boy(str);

the operator= copies the pointer generated in the constructor to t._name. The temporarily created Boy instance goes away, without deleting the internally managed pointer (as it should).

Thus your code works without failure or undefined behavior (but leaks memory).

---

Said all that above, what you really should do, is the following class declaration:

class Boy {
    std::string name;
public:
    Boy(const std::string& n) : name(n) {}
    const std::string& getName() const { return name; }
};

and stop worrying about pointers, memory management and all of that stuff, YAGNI. std::string, and other container classes provided by the c++ standard library are designed to do all of that for you in an efficient, safe, and less error prone manner.