我是C ++编程的新手，我正在阅读这本名为Alex Allain的跳入C ++的书，这本书中有一个井字游戏练习，我很难完成这个练习，所以我所做的就是提示用户输入X或O值，它存储在2d数组中，我想跟踪例如char X是否在数组中出现3次，到目前为止我使用的是Counter ++但它只是增量一旦。下面是我迄今为止所做的事情的来源，希望它能让我的问题更清晰，让我知道我的代码在我的代码的结构和功能方面是如何做的：
#include "stdafx.h"
#include "iostream"
#include "string"
using namespace std;
void display_array(char array[][3]);
void check_input(char array[][3], int input);
void check_winner(char array[][3]);
int check_x(char array[][3]);
int check_o(char array[][3]);
int _tmain(int argc, _TCHAR* argv[])
{
char array[3][3];
int counter = 0;
for(int row = 0; row < 3; row++){
cout << "\n";
for(int col = 0; col < 3; col++){
counter++;
array[row][col] = counter;
}
}
display_array(array);
system("PAUSE");
return 0;
}
void display_array(char array[][3]){
int position_input;
string symbol_input;
do{
for(int i=0; i < 3; i++){
for(int j=0; j < 3; j++){
cout << " [ ";
cout << array[i][j];
cout << " ] ";
}
cout << "\n";
}
cout << "Position: ";
cin >> position_input;
check_input(array, position_input);
}while(position_input != 0);
}
void check_input(char array[][3], int input)
{
char user_input = input;
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
if(user_input == array[i][j])
{
cout << "array[" << i << "][" << j << "] replace with: ";
cin >> array[i][j];
}
}
}
check_winner(array);
}
void check_winner(char array[][3]){
cout << check_x(array);
if(check_x(array) == 3){
cout << check_x(array);
}
else if(check_o(array) == true){
cout << "o";
}
}
int check_x(char array[][3]){
int counter, x[3];
counter = 0;
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
if(array[i][j] == array[i][j]){
counter++;
}
x[i] = counter;
return x[i];
}
}
}
int check_o(char array[][3]){
int counter;
counter = 0;
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
if(array[i][j] == 'o'){
counter++;
return counter;
}else{
return counter;
}
}
}
}
答案 0 :(得分：2)
你有一堆小问题，但最直接的问题似乎就是这样：
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
if(array[i][j] == 'o'){
counter++;
return counter;
}else{
return counter;
}
}
}
无论测试的元素是否为'o'，您都将在一个循环后返回。把它改成这样的东西：
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
if(array[i][j] == 'o')
counter++;
}
}
return counter;
答案 1 :(得分：0)
我的答案：
首先要解决的问题是：编码看起来不错，即使您的程序非常引人注目（// / * * /），如果您对某些操作进行注释，我也会很高兴。
我认为您的问题在解决练习方面有些误导。我一开始以同样的方式想到了你的方式。问题在于模式不仅仅被3的频率检测到，因为它会允许错误的检查模式。使其可视化：
7 8 9
4 5 6
1 2 3
f.e。像1,2,5或9 6 5之类的3的频率不是正确答案，而是-1 5 9。而且，即使您计算出每个频率模式，它的编码也比我在以下解决方案博客中定义的解决方案要多得多：
status_check = possibilities(field,player,2,0,1,1,0,2); //diagonal 1 5 9
status_check = possibilities(field,player,0,0,1,1,2,2); //diagonal 7 5 3
status_check = possibilities(field,player,0,0,1,0,2,0); //vertical 7 4 1
status_check = possibilities(field,player,0,1,1,1,2,1); //vertical 8 5 2
status_check = possibilities(field,player,0,2,1,2,2,2); //vertical 9 6 3
status_check = possibilities(field,player,0,0,0,1,0,2); //horizontal 7 8 9
status_check = possibilities(field,player,1,0,1,1,1,2); //horizontal 4 5 5
status_check = possibilities(field,player,2,0,2,1,2,2); //horizontal 1 2 3
*这就是我认为可以使用数字键盘进行锻炼的方式。（我为那段代码感到非常自豪。:-)）
我很喜欢那本书，但可悲的是我找不到那本书： “奖金：您能否让程序检测到是否无法赢得比赛 在填满整个网格之前在任何一边？” 但是，如果有人跳过了这一章，那么此解决方案肯定会有所帮助：*
练习如下（第121页，跳转到Cpp）：
编写一个小的井字游戏程序，该程序允许两个玩家玩 井字游戏竞争。您的程序应检查是否有 玩家赢了，或者棋盘完全填满了（游戏 以平局结束）。奖励：您能否让您的程序检测游戏是否 整个网格填满之前，哪一方都不能赢？
以下代码可以使用c ++ 17标准进行简单地编译和执行，并且所有库都应随gcc（GCC）8.2.1一起提供
#include <iostream>
#include <string>
#include <limits> //just for the input validation
using namespace std;
int input_check(char field[3][3],char player);
void display_field(char field[3][3],int size);
void initialize_field(char field[3][3],int size,char player);
string win_check(char field[3][3],char player,int size);
int code(char field[3][3],int i,int j,char player);
int possibilities(char field[3][3],char player,int a,int aa,int b,int bb,int c,int cc);
int main()
{
char field[3][3];
int size = 3;
char player ='-';
initialize_field(field,size,player);
display_field(field,size);
while (field[0][0]!='w')
{
player = 'X';
input_check(field,player);
system("clear");
display_field(field,size);
cout << win_check(field,player,size);
if (field[0][0] =='w')
{
cout << endl;
break;
}
player = 'O';
input_check(field,player);
system("clear");
display_field(field,size);
cout << win_check(field,player,size);
if (field[0][0] =='w')
{
cout << endl;
break;
}
}
return 0;
}
void initialize_field(char field[3][3],int size,char player)
{
int i,k;
for (i = 0; i < size; i++)
{
for (k = 0; k < size; k++)
{
field[i][k]= player;
}
}
}
int input_check(char field[3][3],char player)
{
int input = 0;
bool def = false;
while (def != true)
{
while(!(cin >> input))
{ //check the Input format for integer the right way
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "Invalid input. Try again: ";
}
switch (input)
{
case 1:{
def = code(field,2,0,player);
break;}
case 2:{
def = code(field,2,1,player);
break;}
case 3:{
def = code(field,2,2,player);
break;}
case 4:{
def = code(field,1,0,player);
break;}
case 5:{
def = code(field,1,1,player);
break;}
case 6:{
def = code(field,1,2,player);
break;}
case 7:{
def = code(field,0,0,player);
break;}
case 8:{
def = code(field,0,1,player);
break;}
case 9:{
def = code(field,0,2,player);
break;}
default:{
cout << "Invalid input. Try again: " << endl;
break;}
}
}
/*
7 8 9 00 01 02
4 5 6 10 11 12
1 2 3 20 21 22
*/
return input;
}
int code(char field[3][3],int i,int j,char player)
{
int def=0;
if (field[i][j]=='-')
{
field[i][j]=player;
def = true;
}
else
{
cout << "Invalid input. Try again: " << endl;
}
return def;
}
void display_field(char field[3][3],int size)
{
int i,k;
for (i = 0; i < size; i++)
{
for (k = 0; k < size; k++)
{
cout << field[i][k];
cout << " ";
if (k==2) //seperate with new line after the third k
{
cout << endl;
}
}
}
}
string win_check(char field[3][3],char player,int size)
{
string status;
int status_check;
/*
7 8 9 00 01 02
4 5 6 10 11 12
1 2 3 20 21 22
*/
if ((field[0][0]!='-')&&(field[0][1]!='-')&&(field[0][2]!='-') &&(field[1][0]!='-')&&(field[1][1]!='-')&&(field[1][2]!='-')&&(field[2][0]!='-')&&(field[2][1]!='-')&&(field[2][2]!='-'))
{
status = "Rien ne va plus - Nichts geht mehr meine Lieben. Unentschieden";
field[0][0]='w';
}
status_check = possibilities(field,player,2,0,1,1,0,2); //diagonal 1 5 9
status_check = possibilities(field,player,0,0,1,1,2,2); //diagonal 7 5 3
status_check = possibilities(field,player,0,0,1,0,2,0); //vertical 7 4 1
status_check = possibilities(field,player,0,1,1,1,2,1); //vertical 8 5 2
status_check = possibilities(field,player,0,2,1,2,2,2); //vertical 9 6 3
status_check = possibilities(field,player,0,0,0,1,0,2); //horizontal 7 8 9
status_check = possibilities(field,player,1,0,1,1,1,2); //horizontal 4 5 5
status_check = possibilities(field,player,2,0,2,1,2,2); //horizontal 1 2 3
if (status_check == true)
{
status = "Player " + string(1, player) + " won!!";
field[0][0] ='w';
}
return status;
}
int possibilities(char field[3][3],char player,int a,int aa,int b,int bb,int c,int cc)
{
int status;
if ((field[a][aa]==player)&&(field[b][bb]==player)&&(field[c][cc]==player))
{
status = true;
}
return status;
}