无法在构造函数

时间:2016-06-27 19:33:41

标签: c++ oop inheritance

我是C ++的新手。我有简单的Unit类和英雄类,它继承了Unit类。 Hero类有2个额外的参数,但构造函数不能达到父类的参数。 这是unit.hpp:

#ifndef UNIT_HPP
#define UNIT_HPP

#include <string>

using namespace std;

class Unit
{
public:
    unsigned short  max_health  = 100;
    string          name        = "Dummy";
    short           health      = 100;
    short           damage      = 10;
    bool            isDead      = 0;

    Unit();
    Unit(string, unsigned short, unsigned short);
};

#endif //UNIT_HPP

这里是unit.cpp:

#include <string>
#include <iostream>

#include "unit.hpp"

using namespace std;

Unit::Unit()
{
    cout << "Dummy was created!" << endl;
};

Unit::Unit(string N, unsigned short HP, unsigned short AT):
    max_health(HP),
    name(N),
    health(HP),
    damage(AT)
{
    cout << N << " was created!" << endl;
};

这是hero.hpp:

#ifndef HERO_HPP
#define HERO_HPP

#include <string>

#include "unit.hpp"

class Hero : public Unit
{
public:
    unsigned short  max_mana    = 100;
    string          name        = "The Brave Warrior";
    short           mana        = 100;

    Hero (string, unsigned short, unsigned short, unsigned short);

};

#endif //HERO_HPP

最后,这里是hero.cpp:

#include <string>

#include "hero.hpp"

using namespace std;

Hero::Hero(string N, unsigned short HP, unsigned short MP, unsigned short AT):
    max_health(HP),
    max_mana(MP),
    name(N),
    health(HP),
    mana(MP),
    damage(AT)
{
    cout << "The Legendary Hero, " << N << ", was born!" << endl;
}

这是控制台输出:

src/hero.cpp: In constructor ‘Hero::Hero(std::__cxx11::string, short unsigned int, short unsigned int, short unsigned int)’:
src/hero.cpp:10:5: error: class ‘Hero’ does not have any field named ‘max_health’
     max_health(HP),
     ^
src/hero.cpp:13:5: error: class ‘Hero’ does not have any field named ‘health’
     health(HP),
     ^
src/hero.cpp:15:5: error: class ‘Hero’ does not have any field named ‘damage’
     damage(AT)
     ^

问题出在哪里?抱歉英语不好。我希望我提出的问题是正确的,对我来说有这么多新条款。先感谢您。

4 个答案:

答案 0 :(得分:3)

您的基类应负责通常通过构造函数方法初始化其变量。

这:

unsigned short  max_health  = 100;
string          name        = "Dummy";
short           health      = 100;
short           damage      = 10;
bool            isDead      = 0;

看起来不是犹太人。应该在构造函数中初始化这些成员:

Unit::Unit()
: max_health(100),
name("Dummy"),
health(100),
damage(10),
isDead(false)
{ ; }

此外,对于bool个变量,您应该使用truefalse,而不是数字。

编辑1:重复的成员名称
您的子类应避免使用与基类相同的变量名。

英雄行:

  string name;

隐藏或隐藏基类成员:

  string name;

如果您希望保持此约定,则应使用范围解析运算符 ::告诉编译器您指的是哪个成员:

Hero::name = "Hercules"; // Assign member in Hero class
Unit::name = "Person";   // Assign to member in Unit class.   

答案 1 :(得分:0)

在Unit的构造函数中初始化Unit的成员(如果你只想从Hero中调用它,可能是受保护的成员),以及Hero在Hero中的成员。这就是初始化程序列表的工作原理。或者,您可以在Hero的ctor之间使用它们,但不建议这样做。

答案 2 :(得分:0)

C ++不允许你初始化基类&#39;来自子类的成员&#39;初始化列表。

待办事项

Hero::Hero(string N, unsigned short HP, unsigned short MP, unsigned short AT):
    Unit(N, HP, AT), // initializes the base class' members
    max_mana(MP),
    name(N),
    mana(MP),
{
    // but you could override the base class' members here
    isDead = true;
    cout << "The Legendary Hero, " << N << ", was zombified!" << endl;
}

此外,您还有一位名为&#39; name&#39;在Unit和Hero中,你可能不想删除或重命名其中一个。

答案 3 :(得分:0)

您在Hero.cpp构造函数中尝试执行的操作是初始化基类的成员。这对于基类本身来说听起来非常糟糕!实际上,到达分号时,也就是说,基类构造函数Unit()已经被调用(因此,它所包含的内容已经被初始化)。所以,如果你要从Unit中删除Unit(),你会得到一个编译错误,因为另一个构造函数有参数。那些你必须明确指定的那样:

Hero::Hero(string N, unsigned short HP, unsigned short MP, unsigned short AT):
    Unit(HP, N, HP, AT),
    max_mana(MP),
    mana(MP),
    {}

请注意,这次,根本没有调用Unit() - 只有其他的contsructor是 - 显式。