C++函数参数和结构错误

Question

我目前正在编写一个程序，允许用户输入一个方向并沿着坐标平面移动一个机器人。我是C++的新手，所以我得到的一些错误让我感到困惑。有人可以解释我的错误吗？

当我调用函数init_room时，它表示参数太少。当我调用init_robot时会发生同样的事情。他们都有指针参数。我将如何解决这个问题？

当我调用函数在主函数中移动时，它说表达式必须是可修改的值。这是什么意思？

感谢您的帮助！

#include <iostream> 

using namespace std; 

typedef struct room { 
    int x; 
    int y; 
}; 

typedef struct robot { 
    room *current_room; 
    int current_x; 
    int current_y; 
    int model; 
}; 

void init_robot(robot *r) { 
    r->current_room->x; 
    r->current_room->y; 
    r->current_x = 0; 
    r->current_y = 0; 
    //Assign model number 
    r->model; 
} 

void init_room(room *r) { 
    cin >> r->x; 
    cin >> r->y; 

} 

char move(robot *r, int direction) { 
     if (direction == 'N' || direction == 'n') 
     { 
      r->current_y + 1; 
     } 
     else if (direction == 'S' || direction == 's') 
     { 
      r->current_y - 1; 
     } 
     else if (direction == 'E' || direction == 'e') 
     { 
      r->current_x + 1; 
     } 
     else if (direction == 'W' || direction == 'w') 
     { 
      r->current_x - 1; 
     } 
     else 
     { 
      direction = 'x'; 
      return direction; 
     } 

    int main() { 
     char direction; 
     char restart; 

     cout << "What size would you like the room to be?"; 

     room rr; 

     init_room(); 

     robot r; initrobot(); 

     while (true) { 

       cout << "Your robot is in a room with the dimensions (" << rr.x << "," << rr.yy << "). at location (" << r.current_x << "," << r.current_y << ")." << endl; 
       cout << "What direction would you like to move? N (North), E (east), S(South), or W(West)?"; 
       do { 
       cin >> direction; 

       move(direction) = direction; 
       if (direction = 'x') { 
        cout << "Invalid direction. Please enter N, E, S, or W."; 
        } 
       while (direction == 'x'); 

       cout << "Current position is" << endl; 

    //   if (r.current_x <= rr.x && r.current_y <= rr.y) 
     //  { 
      //  cout << "Error, your robot is out of the room bounds. Your robot has exited the room. Would you like to enter another room? Y or N?"; 
        cin >> restart; 
      //} while (restart == 'Y' || restart == 'y'); 
        } 

      system("pause"); 
      return 0; 
     } 

Answer 1

函数init_room和init_robot每个都只有一个参数。所以，你对他们的调用应该像

init_room(&rr); 
init_robot(&r); 

move返回char类型的右值，而右值不能被修改。它只能用作其他表达式的输入。我希望你不知道什么是右值，因此它是一个很好的主意，因为它是C++中的一个重要概念。 （快速版本：一个右值可能在=的右侧，基本上是只读的。）