Question

我在尝试将障碍物移向2x2阵列上的球门时遇到了一些麻烦。我有一个函数，它将两个整数作为参数。然后该函数创建一个大小为10的2x2数组，用自由空间填充它，创建一个设置障碍物（障碍物每次都保持不变）。然后它创建目标点（每次将其分配到相同的点）和球点（在阵列位置[x] [y]中）。

目标是让球向目标移动，直到它撞到障碍物，然后绕过障碍物，同时跟踪障碍物周围每个空间离目标有多远，然后返回到最近点，继续朝着目标前进。

我试图开发一个moveToGoal功能，在没有障碍的情况下将球移向目标，但是我在printGrid函数之外访问网格时遇到了很多麻烦。如果我在打印网格功能之外创建网格，我无法访问它，因为它超出了范围。我知道它可能看起来令人困惑，但我会尝试回答任何有关它的问题。这就是我到目前为止所拥有的：

#include <stdio.h>
#include <vector>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <fstream>
#include <string>
#include <unistd.h>

using namespace std;

void printGrid(int x, int y)        //Used to clear old grids, and print a
{                                   //new grid with input location for ball
    system("CLS");
    string grid [10][10];           //Initialize grid array
    for (int i=0; i<10; i++)
    {
        for (int j=0; j<10; j++)
        {
            grid[i][j] = ' ';       //Constructs grid with freespace
        }
    }
    for (int i=2; i<8; i++)         //Constructs obstacle(s)
    {
        grid[5][i]='O';
    }
    grid[2][5] = 'G';               //Sets Goal location
    grid[x][y] = 'B';               //Sets current ball location(starts 8,5)
    for (int i=0; i<10; i++)        //Prints finished grid
    {
        for (int j=0; j<10; j++)
        {
            cout<<grid[i][j];
        }
        cout<<endl;
    }   
}

void moveToGoal(int x, int y)
{
    printGrid(x-1, y);
}

int main()
{
    moveToGoal(8,5);
    sleep(1);
    moveToGoal(7,5);
    sleep(1);
    moveToGoal(6,5);
    sleep(1);
    moveToGoal(5,5);
    sleep(1);
    moveToGoal(4,5);
    sleep(1);
    moveToGoal(3,5); 
}

任何帮助将不胜感激！

Answer 1

试试这个：

string grid [10][10];           // the grid array is now global

void printGrid(int x, int y)        //Used to clear old grids, and print a
{                                   //new grid with input location for ball
    system("CLS");
    for (int i=0; i<10; i++)  // clean the old grid
    {
        for (int j=0; j<10; j++)
        {
            grid[i][j] = ' ';       //Constructs grid with freespace
        }
    }
    for (int i=2; i<8; i++)         //Constructs obstacle(s)
    {
        grid[5][i]='O';
    }
    grid[2][5] = 'G';               //Sets Goal location
    grid[x][y] = 'B';               //Sets current ball location(starts 8,5)
    for (int i=0; i<10; i++)        //Prints finished grid
    {
        for (int j=0; j<10; j++)
        {
            cout<<grid[i][j];
        }
        cout<<endl;
    }   
}

Answer 2

在以下代码中

int f() {
    int i = 1;
    i++;
    std::cout << i << "\n";
}

int main() {
    f();
    f();
}

函数i中的变量f是局部变量。这意味着每次调用f()时都会创建一个新实例。因此，上述代码的输出为2，两次，不是2，而是3。

在您的代码中，网格是printGrid的本地网格，因此每次调用printGrid时都会创建网格。

解决方案是使网格成为传递给它的参数或全局变量。将网格作为参数传递是首选，您可以考虑将其放在结构或类中以使其更容易。

#include <array>
#include <utility>
#include <iostream>

class Grid
{
    // type alias for a grid of 10x10 chars
    using grid_t = std::array<std::array<char, 10>, 10>;

    grid_t grid_;  // our actual grid data: note grid_[y][x].
    int ballX_ = -1, ballY_ = -1;

public:
    // Constructor
    Grid(int goalX, int goalY)
    {
        // populate the grid with free space
        for (auto&& row: grid_) {
            std::fill(row.begin(), row.end(), ' ');
        }
        // add obstacles
        for (size_t i = 2; i < 8; ++i)
            grid_[i][5] = 'O';
        // place the goal
        grid_[goalY][goalX] = 'G';
    }

    // place, move or remove the ball on the grid
    void placeBall(int x, int y)
    {
        // Remove the ball from any previous location
        if (ballX_ != -1 && ballY != -1)
            grid_[ballY_][ballX_] = ' ';

        // update location
        ballX_ = x, ballY = y;

        // Place ball on board if location is not -1,-1
        if (ballX_ != -1 && ballY != -1)
            grid_[ballY_][ballX_] = 'B';
    }

    // Return the character at a given location
    char at(int x, int y) const
    {
        return grid_[y][x];
    }

    // Query ball position
    int ballX() const { return ballX_; }
    int ballY() const { return ballY_; }

    // print the grid
    friend ostream& operator << (ostream&, const Grid&);
};

ostream& operator<<(ostream& stream, const Grid& grid)
{
    for (auto&& row: grid.grid_) {
        for (auto&& col: row) {
            stream << col << ' ';
        }
        stream << '\n';
    }
    return stream;
}

// helper for your cls/sleep thing around printing the grid.
void printGrid(const Grid& grid)
{
    system("CLS");
    std::cout << grid;
    sleep(1);
}

void manipulateGrid(Grid& grid)  // reference to the caller's grid
{
    printGrid(grid);
    grid.placeBall(8, 5);
    printGrid(grid);
    grid.placeBall(7, 5);
    // ...
}

int main()
{
    Grid g(2, 5);  // create grid with [2,5] as the goal
    manipulateGrid(g);  // to demonstrate passing as parameter
}

如何在c ++

2 个答案: