在定义字符数组的 for 循环内应用输入限制（井字游戏）
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【中文标题】在定义字符数组的 for 循环内应用输入限制（井字游戏）【英文标题】：Applying input limits inside a for loop defining a char array (Tic Tac Toe) 【发布时间】：2020-01-26 07:25:41 【问题描述】：这是我的初始代码 - 用户可以使用它来玩井字游戏，方法是指定他们想要在 char 数组中添加零或交叉的位置 -
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
char a[3][3] = '1', '2', '3', '4', '5', '6', '7', '8', '9';
cout << (char)a[0][0] << " " << (char)a[0][1] << " " << (char)a[0][2] << endl << endl << endl;
cout << (char)a[1][0] << " " << (char)a[1][1] << " " << (char)a[1][2] << endl << endl << endl;
cout << (char)a[2][0] << " " << (char)a[2][1] << " " << (char)a[2][2] << endl << endl << endl;
int x, y;
for (int i = 0; i < 5; i ++)
cout << "Input x: ";
cin >> x;
cout << "Input y: ";
cin >> y;
system("clear");
a[x-1][y-1] = 'X';
cout << (char)a[0][0] << " " << (char)a[0][1] << " " << (char)a[0][2] << endl << endl << endl;
cout << (char)a[1][0] << " " << (char)a[1][1] << " " << (char)a[1][2] << endl << endl << endl;
cout << (char)a[2][0] << " " << (char)a[2][1] << " " << (char)a[2][2] << endl << endl << endl;
cout << "Input x: ";
cin >> x;
cout << "Input y: ";
cin >> y;
system("clear");
a[x-1][y-1] = 'O';
cout << (char)a[0][0] << " " << (char)a[0][1] << " " << (char)a[0][2] << endl << endl << endl;
cout << (char)a[1][0] << " " << (char)a[1][1] << " " << (char)a[1][2] << endl << endl << endl;
cout << (char)a[2][0] << " " << (char)a[2][1] << " " << (char)a[2][2] << endl << endl << endl;
为了避免用户输入[1,3]之外的数组索引，我尝试实现一个if条件
if(x <= 0 && x > 3)
cout << "Wrong! Enter again. Input x: ";
cin >> x;
它不工作。由于 for 循环对 X 和 O 各迭代 5 次，因此输出要么卡在“输入 X”阶段，要么在 5 个“输入 X”后退出。
这是我尝试过的——最终接受了 5 个 X 输入，然后就退出了——
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
char a[3][3] = '1', '2', '3', '4', '5', '6', '7', '8', '9';
cout << (char)a[0][0] << " " << (char)a[0][1] << " " << (char)a[0][2] << endl << endl << endl;
cout << (char)a[1][0] << " " << (char)a[1][1] << " " << (char)a[1][2] << endl << endl << endl;
cout << (char)a[2][0] << " " << (char)a[2][1] << " " << (char)a[2][2] << endl << endl << endl;
int x, y;
for (int i = 0; i < 5; i ++)
cout << "Input x: ";
cin >> x;
if(x <= 0 && x > 3)
cout << "Wrong! Enter again. Input x: ";
cin >> x;
continue;
else
continue;
cout << "Input y: ";
cin >> y;
if(y <= 0 && y > 3)
cout << "Wrong! Enter again. Input y: ";
cin >> y;
continue;
else
continue;
/*
if(x <= 0 && x > 3)
cout << "Wrong! Enter again. Input x: ";
cin >> x;
if(y <= 0 && y > 3)
cout << "Wrong! Enter again. Input y: ";
cin >> y;
*/
system("clear");
a[x-1][y-1] = 'X';
cout << (char)a[0][0] << " " << (char)a[0][1] << " " << (char)a[0][2] << endl << endl << endl;
cout << (char)a[1][0] << " " << (char)a[1][1] << " " << (char)a[1][2] << endl << endl << endl;
cout << (char)a[2][0] << " " << (char)a[2][1] << " " << (char)a[2][2] << endl << endl << endl;
cout << "Input x: ";
cin >> x;
if(x <= 0 && x > 3)
cout << "Wrong! Enter again. Input x: ";
cin >> x;
continue;
else
continue;
cout << "Input y: ";
cin >> y;
if(y <= 0 && y > 3)
cout << "Wrong! Enter again. Input y: ";
cin >> y;
continue;
else
continue;
/*
if(x <= 0 && x > 3)
cout << "Wrong! Enter again. Input x: ";
cin >> x;
if(y <= 0 && y > 3)
cout << "Wrong! Enter again. Input y: ";
cin >> y;
*/
system("clear");
a[x-1][y-1] = 'O';
cout << (char)a[0][0] << " " << (char)a[0][1] << " " << (char)a[0][2] << endl << endl << endl;
cout << (char)a[1][0] << " " << (char)a[1][1] << " " << (char)a[1][2] << endl << endl << endl;
cout << (char)a[2][0] << " " << (char)a[2][1] << " " << (char)a[2][2] << endl << endl << endl;
为什么我不能只在输入后有一个 if 语句，其中它检查它是否超出范围，如果是，它可以再次询问用户，然后继续？ - 这是我想要实现的。
【问题讨论】：
提示：for loop 正在迭代 5 次，无论用户输入是否错误。您需要继续游戏，直到一位用户success 或所有瓷砖都被填满。甚至你可以为输入添加一个循环循环，直到它有效为止。
【参考方案1】：
我认为正确的运算符是 ||而不是 &&。
通过使用 &&，您建议必须同时发生 x3 才能采用 IF 语句！
尝试用英语大声朗读你的陈述会有所帮助，有助于你更好地理解它:)
【讨论】：
【参考方案2】：为什么不能只在输入后有一个if语句，检查是否越界，如果是，可以再次询问用户，然后继续？
if 语句检查用户输入的有效性，但在输入有效之前它不会迭代。您需要有一个loop，它会一直运行到用户输入有效为止。
如下所示，
#include <iostream>
bool isGameCompleat(char a[3][3])
/*check for all combination for the board for game complete*/
if ((a[0][0] == a[0][1] && a[0][0] == a[0][2]) ||
(a[1][0] == a[1][1] && a[1][0] == a[1][2]) ||
(a[2][0] == a[2][1] && a[2][0] == a[2][2]) ||
(a[0][0] == a[1][0] && a[0][0] == a[2][0]) ||
(a[0][1] == a[1][1] && a[0][1] == a[2][1]) ||
(a[0][3] == a[1][3] && a[0][3] == a[2][3]) ||
(a[0][0] == a[1][1] && a[0][0] == a[2][2]) ||
(a[0][3] == a[1][1] && a[0][3] == a[2][0]))
return true;
return false;
void printBoard(char a[3][3])
std::cout << (char)a[0][0] << " " << (char)a[0][1] << " " << (char)a[0][2] << "\n\n\n";
std::cout << (char)a[1][0] << " " << (char)a[1][1] << " " << (char)a[1][2] << "\n\n\n";
std::cout << (char)a[2][0] << " " << (char)a[2][1] << " " << (char)a[2][2] << "\n\n\n";
int main()
char a[3][3] = '1', '2', '3', '4', '5', '6', '7', '8', '9' ;
int x, y;
printBoard(a);
for (int i = 0; (i < 9); i++)
int p;
if ((i & 1) == 0)
std::cout << "X Turn ";
else
std::cout << "O Turn ";
std::cout << "Enter the position ";
std::cin >> p;
/*derived row and column from 0 to 9 position*/
x = (p / 3);
y = (p%3) -1;
/*Validity check for input if wrong request for new input
* out of range check and non-repeated tile is selected
*/
while (p < 0 || p > 9 || (a[x][y] =='X') || (a[x][y]=='O'))
std::cout << "Wrong! Enter again: ";
std::cin >> p;
x = (p / 3);
y = (p%3) -1;
if ((i & 1) == 0)
a[x][y] = 'X';
else
a[x][y] = 'O';
system("clear");
printBoard(a);
if (isGameCompleat(a))
std::cout <<( (i & 1) ? "O" : "X") << " wins";
break;
else if (i == 8)
std::cout << "Game is draw";
【讨论】：
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