是否可以在C ++中从构造方法访问对象名称？

杰克逊123

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;
}

因此，基本上，如果您看一下cout构造函数末尾的函数，我希望语句以对象名称开头而无需硬编码。因此，例如，一个对象被命名为TylerI，我希望第一条cout语句打印出来Tyler's Health is XYZ。

我希望这项工作能够使我能够创建一个对象，而不必每次都对名称进行硬编码。

如果这是对我要达到的目标的不好解释，我感到抱歉。预先感谢您提供的任何帮助！

ΦXocę웃Пepeúpatsu

这种解释或反射在C ++中很难

最好的办法是修改类，并为每个对象指定一个属性名称：

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

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;
}

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