Question

我正在创建一个程序，用户可以在打印到屏幕的数组中移动他们的角色（数字1）。当我试图检查下一个位置是否在我的moveRight函数中打开时，我遇到了麻烦。

我想返回数组部分的值，这是1的右边一个空格。我试图返回数组下一个点的值的原因是因为我想返回该值我的drawBoard函数，所以我可以使用该值重新打印板，使一个在该位置。我如何将mB [i + 1] - （1的右边的下一个值）返回到我的drawBoard函数？

#include "stdafx.h"
#include <iostream>
#include "PlayerM.h"

using namespace std;
class Player {
public:
   Player::Player(int b[]) //create a constructer to copy the values of b into mB
   {
      // copy b into the member array
      for (int i = 0; i < 16; i++) {
         mB[i] = b[i];
      }
   }

   int moveUp();
   void moveDown();
   int moveRight();
   void moveLeft();
private:
   int mB[16];

};
int Player::moveUp() {
   return 0;
}
void Player::moveDown() {

}
int Player::moveRight() {
   for (int i = 0; i < 16; i++) //find the players pos on aray
   {
      if (mB[i] == 1 && i < 3) //if the player is eligible to move
      {
         mB[i] = 0;
         mB[i + 1] = 1;
         return mB[i + 1];
      }
   }
}
void Player::moveLeft() {
}
int drawBoard(int boardArray[16]) //draw the game board
{
   for (int i = 0; i < 16; i++) //use a for loop to simply draw the game board (4x4)
   {
      cout << boardArray[i]; //ouput the storage id of the array
      if (i == 3 || i == 7 || i == 11 || i == 15) //every 4 lines begin new line
      {
         cout << "\n";
      }
   }
   return 0;
}
int main() {
   int bArray[16] = {0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}; //create an array [16]
   drawBoard(bArray); //send the aray to drawBoard ()
   Player p(bArray); //send bArray to the p constructer
   char m;
   cin >> m;

   if (m == 'W') {
      p.moveRight();
   }

   char f;
   cin >> f;
}

Answer 1

int Player::moveRight() {
   for (int i = 0; i < 16; i++) //find the players pos on aray
   {
      if (mB[i] == 1 && i < 3) //if the player is eligible to move
      {
         mB[i] = 0;
         mB[i + 1] = 1;
         return mB[i + 1];
      }
   }
}

我认为您需要以下内容（假设主板为4x4）：

int Player::moveRight() {
   for (int i = 0; i < 16; i++) //find the players pos on aray
   {
      if (mB[i] == 1 && (i%4) < 3) //if the player is eligible to move
      {
         mB[i] = 0;
         mB[i + 1] = 1;
         return mB[i + 1];
      }
   }
}

更改内容：i < 3至(i%4) < 3

将值从函数返回到另一个函数

1 个答案: