c++ - 是否可以从 C++ 中的构造函数访问对象名称？

Question

所以我的问题是是否可以从构造函数访问正在构造的对象的名称。这是我的代码片段:

Monk::Monk(int stam, int agil, string spec){
    stamina = stam;
    agility = agil;
    specialization = spec;
    cout << "'s Health is " << health() << endl;
    cout << "'s DPS is " << damage() << endl;
    cout << "'s current specification is a " << specName() << " monk." << endl;
}
int main() {
    Monk Tyler(25000, 1245, "Brewmaster");
    Monk Jackson(12500, 3000, "Windwalker");
    return 0;
}

所以基本上，如果您查看构造函数末尾的 cout 函数，我希望语句以对象的名称开头，而无需硬编码。因此，例如，其中一个对象名为 Tyler 我希望第一个 cout 语句打印出 Tyler's Health is XYZ。

我希望它能工作，这样我就可以创建一个对象，而不必每次都对名称进行硬编码。

如果这是对我想要实现的目标的错误解释，我很抱歉。提前感谢您提供的任何帮助!

Answer 1

那种instrocpection或反射在C++中有点难

你能做的最好的就是修改类并给每个对象一个属性名:

Monk::Monk(int stam, int agil, string spec, string name){
    stamina = stam;
    agility = agil;
    specialization = spec;
    monkName = name;
    cout << monkName << "'s Health is " << health() << endl;
    cout << monkName << "'s DPS is " << damage() << endl;
    cout << monkName << "'s current specification is a " << specName() << " monk." << endl;
}

int main() {
    Monk Tyler(25000, 1245, "Brewmaster");
    Monk Jackson(12500, 3000, "Windwalker");
    return 0;
}

---

旁注，记住你可以做到

Monk::Monk(int stam, int agil, string spec, string name):stamina(stam),agility (agil),specialization(spec),monkName(name){
    //stamina = stam;
    //agility = agil;
    //specialization = spec;
    //monkName = name;
    cout << monkName << "'s Health is " << health() << endl;
    cout << monkName << "'s DPS is " << damage() << endl;
    cout << monkName << "'s current specification is a " << specName() << " monk." << endl;
}