C++中具有基类和派生类的作用域

Scope with Base and Derived Classes in C++

本文关键字：派生 作用域 基类 C++      更新时间：2023-10-16

我很难理解我在这里遇到了什么问题。这是学校一门课程的作业。我在笔记本电脑上写代码，并在学校的服务器上编译/测试/提交。

我目前在clion中编写代码。当我在Mac的终端上运行gcc -v或g++ -v时，我会得到以下信息：

Configured with: --prefix=/Library/Developer/CommandLineTools/usr --with-gxx-include-dir=/usr/include/c++/4.2.1
Apple LLVM version 7.0.0 (clang-700.1.76)
Target: x86_64-apple-darwin15.0.0

在学校的服务器上运行相同的命令我得到：

gcc version 4.4.7 20120313 (Red Hat 4.4.7-16) (GCC)

我在不同版本的gcc上编译，不确定这是否会影响我的问题。向前

int main() {
    int choice; // Used to get creature selection from user
    Creature *creature1, *creature2; // Objects created
    printCreatureList(); // Prints list of creatures for players to select from
    choice = getIntFromUser(5); // Gets user choice for creature selection
    if (choice == 1) {
        Goblin newGob1;
        creature1 = &newGob1;
        newGob1.setStats();
        cout << "Created " << creature1->getName() << " as player 1's creature.n";
// more if-else, and repeat for player 2 ...
    }

现在，玩家1和玩家2各自创建了一个准备战斗的生物。注意，为了以后使用，creature1->getName()在此处正常工作。以下是战斗循环中给我带来问题的部分。请注意，还有另一个版本，玩家2进攻，玩家1防守。

    do { // Enter game loop
        cout << endl << "nTurn #" << i << ", Player 1 (" << creature1->getName() << ") attacking Player 2 (" << creature2->getName() << ")";
        i++;
        p1Attacks(*creature1, *creature2, *p1Achilles, *p2Achilles); // Player 1 attacks, player 2 defends
        if (creature2->getStrength() <= 0) { //Check if creature2 was defeated
            cout << "nt***Player 2's creature has taken fatal damage***" << endl;
            cout << "nt* * * Player 1 (" << creature1->getName() << ") has won the battle * * *" << endl;
            winCondition = false;
            break;
       // advances on to p2Attacks
     } while (winCondition);

我的p1Attack和p2Attack有类似的格式：

void p1Attacks(Creature &p1, Creature &p2, bool &p1AchillesInjury, bool &p2AchillesInjury)

p1攻击/p2攻击工作正常，所有的计算结果都很完美。但当我在学校服务器上运行战斗时，gcc 4.4.7 20120313我看到：

Turn #1, Player 1 () attacking Player 2 () Player 1's attack roll: 7 Player 2's defend roll: 1 Player 1's damage output: 6 Player 2's armor: 3 Player 2 damage taken: 3 Player 2 new strength: 5

第一行不正确，如果他们每个人都创建了相应的字符，那么应该读Turn #1, Player 1 (The Barbarian) attacking Player 2 (Reptile)。在我的本地机器上，代码运行正确，并正确拼写括号中的名称。

我的生物类和creature.cpp:中的.setStats（）示例

class Creature {
public:
    Creature() {}
/*
    Functions:      changeStrength()
    Description:    Change strength attribute for creature by reducing value
    Parameters:     reduceStrengthBy
    Preconditions:  None
    Postconditions: Strength is reduced
*/
    void changeStrength(int reduceStrengthBy);
    int getAttackDice()     { return attackDice; }
    int getAttackSides()    { return attackSides; }
    int getDefenseDice()    { return defenseDice; }
    int getDefenseSides()   { return defenseSides; }
    int getArmor()          { return armor; }
    int getStrength()       { return strength; }
    std::string getName()   { return name; }
    bool getDodge()         { return dodge; }
protected:
    int attackDice;
    int attackSides;
    int defenseDice;
    int defenseSides;
    int armor;
    int strength;
    std::string name;
    bool dodge;
};
void Reptile::setStats() {
    attackDice = 3;
    attackSides = 6;
    defenseDice = 1;
    defenseSides = 6;
    armor = 7;
    strength = 18;
    name = "The Reptile";
    dodge = false;
}

因此，我的最终问题是，为什么在早期的if statement中，线路creature1->getName()在我的笔记本电脑和学校服务器上都能正常工作，但只在我的本地机器上工作，以后无法在远程服务器上工作（接近p1Attack）？

如果某个东西在一台机器上工作，但在另一台机器不工作，那么您可以确定自己已经成为未定义行为的受害者。

事实上，你在使用悬空指针。基本上，它可以归结为：

int main(int, char**) {
  int * pointer; // uninitialised, don't use
  if (someCondition) {
    int object = 42;
    pointer = &object; // assigned to point to an object, can be used
  } // object goes out of scope here
  // pointer is dangling, don't use
  cout << *pointer << endl; // oops
  return 0;
}

要解决这个问题，您需要将对象存储在不受自动内存管理的位置（例如堆上），或者重新排列代码以利用它。

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