函数返回空白

huangapple
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英文:

Function returning blank

问题

我在尝试测试C++中非常基本的面向对象编程，但遇到了一个问题。我正在编写以下类：

演员类：

  1. #pragma once
  2. #include <string>
  3. class Actor
  4. {
  5. private:
  6. 	void setName(std::string inputName);
  8. public:
  9. 	std::string name{};
  11. 	Actor(std::string inputName = "Actor");
  13. 	std::string getName();
  15. };
  1. #include "Actor.h"
  3. Actor::Actor(std::string inputName)
  4. {
  5. 	setName(inputName);
  6. }
  8. void Actor::setName(std::string inputName)
  9. {
  10. 	name = inputName;
  11. }
  13. std::string Actor::getName()
  14. {
  15. 	return name;
  16. }

角色类

  1. #pragma once
  2. #include "Actor.h"
  4. class Character : public Actor
  5. {
  6. public:
  7. 	int maxHealth{};
  8. 	int currentHealth{};
  10. 	Character(std::string inputName = "Character", int inputHealth = 100);
  11. 	int getCurrentHealth();
  13. };
  1. #include "Character.h"
  3. Character::Character(std::string inputName, int inputHealth) :
  4. 	Actor(inputName), maxHealth(inputHealth), currentHealth(inputHealth)
  5. {
  7. }
  9. int Character::getCurrentHealth()
  10. {
  11. 	return currentHealth;
  12. }

玩家类

  1. #pragma once
  2. #include "Character.h"
  3. class Player : public Character
  4. {
  5. public:
  6. 	Player();
  8. };
  1. #include "Player.h"
  3. Player::Player() :
  4. 	Character("Player", 200)
  5. {
  7. }
  1. #include <iostream>
  2. #include "Player.h"
  4. int main()
  5. {
  6. 	Player a{};
  8. 	std::cout << "Actor name is: " + a.getName() << std::endl;
  9. 	std::cout << "Player health is: " + a.getCurrentHealth() << std::endl;
  11. 	return 0;
  12. }

在这种情况下，a.getName() 正常打印，返回字符串 "Player"。然而，a.getCurrentHealth 没有打印。不仅仅是值，而是整个句子 "Player health is: " 都没有打印。有关发生了什么的任何想法吗？

英文:

I was trying to test very basic OO in c++, but I ran into a problem. I was writing the following classes:

Actor Class:

  1. #pragma once
  2. #include &lt;string&gt;
  3. class Actor
  4. {
  5. private:
  6. 	void setName(std::string inputName);
  8. public:
  9. 	std::string name{};
  10. 	
  11. 	Actor(std::string inputName = &quot;Actor&quot;);
  13. 	std::string getName();
  15. };
  1. #include &quot;Actor.h&quot;
  3. Actor::Actor(std::string inputName)
  4. {
  5. 	setName(inputName);
  6. }
  8. void Actor::setName(std::string inputName)
  9. {
  10. 	name = inputName;
  11. }
  13. std::string Actor::getName()
  14. {
  15. 	return name;
  16. }

Character Class

  1. #pragma once
  2. #include &quot;Actor.h&quot;
  4. class Character : public Actor
  5. {
  6. public:
  7. 	int maxHealth{};
  8. 	int currentHealth{};
  10. 	Character(std::string inputName = &quot;Character&quot;, int inputHealth = 100);
  11. 	int getCurrentHealth();
  13. };
  1. #include &quot;Character.h&quot;
  3. Character::Character(std::string inputName, int inputHealth) : 
  4. 	Actor{inputName}, maxHealth{inputHealth}, currentHealth{inputHealth}
  5. {
  7. }
  9. int Character::getCurrentHealth()
  10. {
  11. 	return currentHealth;
  12. }

Player Class

  1. #pragma once
  2. #include &quot;Character.h&quot;
  3. class Player : public Character
  4. {
  5. public:
  6. 	Player();
  8. };
  1. #include &quot;Player.h&quot;
  3. Player::Player():
  4. 	Character{&quot;Player&quot;, 200}
  5. {
  6. 	
  7. }
  1. #include &lt;iostream&gt;
  2. #include &quot;Player.h&quot;
  4. int main()
  5. {
  6. 	Player a{};
  8. 	std::cout &lt;&lt; &quot;Actor name is: &quot; + a.getName() &lt;&lt; std::endl;
  9. 	std::cout &lt;&lt; &quot;Player health is: &quot; + a.getCurrentHealth() &lt;&lt; std::endl;
  10. 	
  12. 	return 0;
  13. }

In this case, a.getName() prints normally, returning the string "Player". However, a.getCurrentHealth does not print. Not just the value, but the entire sentence "Player health is: " is not printing. Any idea on what's happening?

答案1

得分: 5

C++中没有将整数添加到字符串的操作。使用 << 而不是 + 来获得您想要的输出。

  1. std::cout << "Player health is: " << a.getCurrentHealth() << std::endl;

您的代码正在执行 指针算术。字符串文字被解释为 char 指针，然后将健康值添加到该指针，然后将结果地址（可能指向任何地方）打印出来，就好像它指向了C样式字符串。

英文:

There's no operation in C++ that adds an integer to a string. Use &lt;&lt; not + to get the output you want.

  1. std::cout &lt;&lt; &quot;Player health is: &quot; &lt;&lt; a.getCurrentHealth() &lt;&lt; std::endl;

What your code is doing is pointer arithmetic. The string literal was interpreted as a char pointer, the health value was then added to that pointer and the resulting address (which could have been pointing anywhere) was printed as if it pointed to a C style string.

huangapple
  • 本文由 发表于 2023年8月4日 20:58:26
  • 转载请务必保留本文链接：https://go.coder-hub.com/76836144.html
  • c++
  • oop
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