c++ - 海龟图形程序

Question

#include <iostream>
#include <fstream>
using namespace std;

void printArray(void);  
char grid[6][34] = {'0'}; // empty floor
void movement(int command);
int direction = 2; // 1 for north, 2 for east, 3 for south, 4 for west. Turtle starts facing east.
void turtleMovement(int, int);
int spacesMoved; // used after F command.
int yCoord = 0; // Y position assumed 0;
int xCoord = 0; // X position assumed 0;
bool pen = 0; // assume pen is up
void initialize(void);
char command; //  U for pen up, D for pen Down, R for right turn, L for left turn, F for forward movement with spaces moved, P for array print, and Q for end.
int main(){

    initialize();
    ifstream infile;
    infile.open("commands.txt");
    infile >> command;
    while(command != 'Q'){
        if(command == 'U'){
            pen = 0;
        }
        else if(command == 'D'){
            pen = 1;
        }
        if(command == 'R' || command == 'L'){
            movement(command);
        }

        if(command == 'F'){
            infile >> spacesMoved;
            turtleMovement(yCoord, xCoord);
        }
        else if(command == 'P')
            printArray();
        infile >> command;
        }
}
//=====================================================================================================================================================
//
// Function Name: intialize
//
// Pre: none
//
// Post: sets every element of the array grid to all zeroes.
// 
//======================================================================================================================================================
void initialize()
{
    for(int i = 0; i < 6; i++){     //
        for(int j = 0; j < 34; j++){    // Initialization of Array to all zeros.
            grid[i][j] = '0';           //
        }
    }
}
//=====================================================================================================================================================
//
// Function Name: printArray
//
// Pre: none
//
// Post: prints out the array grid to the console
// 
//======================================================================================================================================================
void printArray()
{
    for(int i = 0; i < 6; i++)
                for(int j = 0; j < 34; j++){    
                    grid[i][j];
                    cout << grid[i][j] << (j == 33 ? "\n" : ""); // if the row gets to 33 characters a new line will be made.
                }   
}
void movement(int command)
{
    if (direction == 2){ // direction is east
        if (command == 'R')
            direction = 3;// if you're facing east and you turn right you're now facing south 
        else if (command == 'L') // if you're facing east and you turn left you're now facing north.
            direction = 1;
    }
        else if (direction == 3) // direction is south
    {
        if (command == 'R') // turn right to go west
            direction = 4;
        else if (command == 'L') // turn left to go east
            direction = 2;
    }
    else if (direction == 4) // direction is west
    {
        if (command == 'R') // right for north
            direction = 1;
        else if (command == 'L') // left for south
            direction = 3;
    }
    else if (direction == 1) // direction is north
    {
        if (command == 'R') // right for east 
            direction = 2;
        else if (command == 'L') // left for west
            direction = 4;
    }
}
void turtleMovement(int y, int x){
    if(direction == 2){
        int i = y;
                for(int j = x; j < x + spacesMoved; j++){
                    if(pen == 1)
                        grid[i][j] = '1';
                    else
                        grid[i][j] = '0';
                }
            xCoord += spacesMoved - 1;   //updates the x coordinate
    }           


        else if(direction == 3){
            for(int i = y; i < y + spacesMoved; i++){
                int j = x;
                    if(pen == 1)
                        grid[i][j] = '1';
                    else
                        grid[i][j] = '0';

            }
            yCoord += spacesMoved - 1;          //updates the y coordinate
        }
        else if(direction == 4){
            int i = y;
                for(int j = x; j > x - spacesMoved; j++){
                    if(pen == 1)
                        grid[i][j] = '1';
                    else if (pen == 0)
                        grid[i][j] = '0';
                }
            xCoord -= spacesMoved - 1;         //updates the x coordinate
        }
        else if(direction == 1){
            for(int i = y; i > y- spacesMoved; i++){
                int j = x;
                    if(pen == 1)
                        grid[i][j] = '1';
                    else
                        grid[i][j] = '0';
            }
            yCoord -=spacesMoved - 1;       //updates the y coordinate
        }
}

我对我的程序感到极度沮丧，当我尝试进行 180 度转弯时，即 R R 或 L L，整个程序就会卡住。跟我的逻辑有关系吗？还是语法？因为它不会给我任何这些错误。

Answer 1

抱歉，您的程序让我很烦，而且有点难读。

试试这个:

struct Position
{
  int x;
  int y;
};

struct Direction_Vector : Position
{
  unsigned int direction;
};

const unsigned int MAX_DIRS = 4u;


void movement(char command,
              const Direction_Vector present_dir,
              Direction_Vector&       new_dir)
{
  unsigned int dir = present_dir.direction; 
  switch (command)
  {
    case 'R':  // rotate clockwise;
      dir = (dir + 1) % MAX_DIRS;
      break;
    case 'L':  // rotate counter clockwise
      dir = (dir + (MAX_DIRS - 1)) % MAX_DIRS;
      break;
    case 'M':  // Move one position in the given direction
      switch (dir)
      {
        case 0:  // Assume north, Y decreases
           new_dir.y = present_dir.y - 1;
           break;
        case 1:  // Assume east, X increases
           new_dir.x = present_dir.x + 1;
           break;
        case 2:  // Assume south, Y increases
           new_dir.y = present_dir.y - 1;
           break;
        case 3:  // Assume west, X decreases
           new_dir.x = present_dir.x - 1;
           break;
      }
      break;
    default:
      break;
  }
  new_dir.direction = dir;
}

您可以通过使用代数并实现方向和调整后的纵坐标之间的关系来优化运动部分。

此外，我还没有进行任何边界检查。

如果我简化了您发布的大量代码而毁了您的一天，我深表歉意。

使用调试器单步执行上述代码，看看它是否正确处理了 180 度转弯。

编辑 1:真值表
这是运动和纵坐标变化的真值表

Direction | Delta X | Delta Y|  // "delta" means change  
----------+---------+--------+  
    0     |    0    |   -1   |  
----------+---------+--------+  
    1     |    +1   |    0   |  
----------+---------+--------+  
    2     |    0    |   +1   |  
----------+---------+--------+  
    3     |    -1   |    0   |  
----------+---------+--------+  

你能想出一个基于方向的纵坐标代数方程吗？