C++中的基类和派生类的范围

Question

我很努力地理解我在这里遇到的问题。这是学校课程的一项任务。我在笔记本电脑上编写我的代码，并在学校的服务器上编译/测试/提交。

我目前在clion中写我的代码。当我在我的Mac的终端上运行gcc -v或g++ -v我得到如下：

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 

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) 

我编写不同版本的gcc，不知道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各自拥有一个可以战斗的生物。请注意以后使用，creature1->getName()在此处正确运行。这里是战斗循环的一部分，给我一个问题。请注意，还有另外一个版本，玩家2攻击，玩家1进行防御。

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 << "\n\t***Player 2's creature has taken fatal damage***" << endl; 
      cout << "\n\t* * * Player 1 (" << creature1->getName() << ") has won the battle * * *" << endl; 
      winCondition = false; 
      break; 

     // advances on to p2Attacks 
    } while (winCondition); 

我p1Attack和p2Attack也有类似的格式：

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

的p1Attacks/p2Attacks正常工作，并且所有的数学出来完美。但是，当我在我的学校服务器上运行的斗争中，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

第一行是不正确，应该读Turn #1, Player 1 (The Barbarian) attacking Player 2 (Reptile)如果他们每个创建的那些各自的特点。在我的本地机器上，代码运行正确，并按照它应该用圆括号拼写出来。

我的生物类和.setStats（）从creature.cpp的例子：

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; 
} 

所以我最终的问题是，为什么正确地在我的笔记本电脑，而在学校的服务器都行creature1->getName()功能早期的if statement，但只能在我的本地机器上工作，并且以后无法在远程服务器上工作（靠近p1Attack）？

Answer 1

如果在一台机器上有效但不是另一台机器，则可以确定您已成为未定义行为的受害者。

事实上，你使用的是悬挂指针。基本上，它归结为：

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; 
} 

为了解决这个问题，你需要可以存储你的对象，他们是不受自动内存管理（例如，在堆）或重新安排你的代码，以利用它的优势。