C++ 我无法访问我需要的功能

Question

我正在尝试用 C++ 为我的学校项目创建一个基于文本的冒险游戏。我遇到的问题是我的 gameover() 函数需要能够转到我的 begin() 函数。问题是必须在 gameover() 函数之前声明 begin 以允许它转到 begin()，只有我有其他函数也需要访问 gameover(。简而言之，我需要一种方法来告诉我的程序转到函数 gameover() 或 begin()，并知道它存在并已声明。谢谢，西蒙

void begin() {
   int name;
   int choice1;
   system("cls");
   cout << "To start your adventure, please enter your name." << endl;
   cin >> name;
   cin.ignore();
   system("cls");
   cout << "Hello " << name << " Your adventure now begins.... Who knows what's in store for you!" << endl;
   system("pause");
   system("cls");
   cout << "You find yourself in a dark, cramp library. " << endl;
   cout << "You don't know how you got here, or where you are." << endl;
   cout << "Luckily there is a sword laying on the ground next to you \nand an open door in front.\n" << endl;
   cout << "What will you do?" << endl;
   cout << "1. Pick up the sword" << endl;
   cout << "2. Walk forward through the door" << endl;
   cout << "3. Use the sword to end your miserable existence!" << endl;
   cin >> choice1;
   cin.ignore();
   system("cls");
   if (choice1 == 1) {
      cout << "You quickly pick up the sword and run through the door." << endl;
      system("pause");
      road();
   }
   else if (choice1 == 2) {
      cout << "While you make you way to the door...." << endl;
      cout << "You somehow managed to trip on the sword." << endl;
      cout << "You fall on the hilt smashing your neck, and end your painfully short life. " << endl;
      system("pause");
      gameover();
   }
   else   {
      cout << "That wasn't an option....." << endl;
      cout << "You have now broken the game. Good day Sir!!!" << endl;
   }
 }


 void gameover() {
    int choice_b;
    cout << " Oops! You died.... Try Again." << endl;
    cout << "\n1. Start Again!" << endl;
    cout << "2. Exit" << endl; 

    cin >> choice_b;
    cin.ignore();
    system("cls"); 

    if (choice_b == 1) {
        begin();
    }
    else if (choice_b == 2) { std::exit; }
 }

Answer 1

C++ 要求您在调用语句之前描述函数。如果您要在 main() 顶部添加定义，那么它的调用语句将在任何地方工作，第二个选项是在调用之前声明函数。由您决定要在何处访问该功能。

基本上，如果您在头文件或 main 顶部添加一些声明，那么这些函数将在任何地方运行

#include headerfiles....
void begin(); //
void end(); // Function prototypes 
int main()
{
   .....
    begin(); // Will work here
}