【发布时间】：2015-04-24 05:53:38
【问题描述】：
我一直在转来转去，试图弄清楚如何让我的球员轮流。问题是有两个玩家在玩井字游戏，但我不知道该怎么做。这是我的代码：
#include <stdio.h>
#include <stdlib.h>
void displayBoard(char [3][3]);enter code here
int playerType (int player, char boardArray[3][3]);
int selectLocation(char [3][3], int , int );
char setTurn(char [3][3], int , int , char );
int main()
{
int player,location;
char position;
char boardArray[3][3]={{'1','2','3'},
{'4','5','6'},
{'7','8','9'}};
player= playerType (player, boardArray);
int i;
for(i=0;i<9;i++)
{
if (player==3)
break;
else{
location=selectLocation(boardArray, player, location);
position=setTurn(boardArray, location, player, position);
}
}
return 0;
}
void displayBoard(char boardArray [3][3]) //This displays the tic tac toe board
{
printf("\t%c|%c|%c\n", boardArray[0][0], boardArray[0][1], boardArray[0][2]);
printf("\t%c|%c|%c\n", boardArray[1][0], boardArray[1][1], boardArray[1][2]);
printf("\t%c|%c|%c\n", boardArray[2][0], boardArray[2][1], boardArray[2][2]);
}
int playerType (int player, char boardArray [3][3]) //This decides who plays first
{
player=0;
printf("Enter 1 for Player X.\n");
printf("Enter 2 for Player O.\n");
printf("Enter 3 to Quit. \n");
scanf("%d", &player);
if (player == 1)
{
printf("You're player X.\n");
displayBoard(boardArray);
}
else if (player == 2)
{
printf("You're player O.\n");
displayBoard(boardArray);
}
else if(player == 3)
printf("You Quit.\n");
else
printf("Invalid Entry.\n");
return player;
}
int selectLocation(char boardArray[3][3], int player, int location) //This takes in the location
{
printf("Pick a location from 1-9.\n");
scanf("%d", &location);
return location;
}
char setTurn(char boardArray[3][3], int location, int player, char position) //This outputs the location
{
if (player == 1)
{
switch(location)
{
case 1:
{
boardArray[0][0]='x';
break;
}
case 2:
{
boardArray[0][1]='x';
break;
}
case 3:
{
boardArray[0][2]='x';
break;
}
case 4:
{
boardArray[1][0]='x';
break;
}
case 5:
{
boardArray[1][1]='x';
break;
}
case 6:
{
boardArray[1][2]='x';
break;
}
case 7:
{
boardArray[2][0]='x';
break;
}
case 8:
{
boardArray[2][1]='x';
break;
}
case 9:
{
boardArray[2][2]='x';
break;
}
default:
printf("invalid");
}
}
else if (player == 2)
{
switch(location)
{
case 1:
{
boardArray[0][0]='O';
break;
}
case 2:
{
boardArray[0][1]='O';
break;
}
case 3:
{
boardArray[0][2]='O';
break;
}
case 4:
{
boardArray[1][0]='O';
break;
}
case 5:
{
boardArray[1][1]='O';
break;
}
case 6:
{
boardArray[1][2]='O';
break;
}
case 7:
{
boardArray[2][0]='O';
break;
}
case 8:
{
boardArray[2][1]='O';
break;
}
case 9:
{
boardArray[2][2]='O';
break;
}
default:
printf("Invalid");
}
}
printf("\t%c|%c|%c\n", boardArray[0][0], boardArray[0][1], boardArray[0][2]);
printf("\t%c|%c|%c\n", boardArray[1][0], boardArray[1][1], boardArray[1][2]);
printf("\t%c|%c|%c\n", boardArray[2][0], boardArray[2][1], boardArray[2][2]);
return position;
}
【问题讨论】：
-
那么......你被困在哪里了？什么是预期的 o/p 和什么是实际的？
-
我宁愿将 boardArray 初始化为全 0，并将 boardindex 设置为 1 或 2 以指示哪些玩家拥有该索引。但回到你的问题......这似乎不清楚......你的开关的哪个功能？还是转弯功能还有待创建？
-
转弯功能仍有待创建。我在想也许我会有一个嵌套的 for 循环，也许？