如何在if语句中比较二维数组3个字符元素？

问题描述

在第 20、27、32、36 行中，即使该值为真，计数值在最后并未增加，但显示为 0 计数值。我想我在比较数组时犯了一个错误。我不知道如何比较字符元素二维数组 输入是
1
xxx
哦

---

答案必须是 3 代码如下

#include <stdio.h>
    int main()
    {
        int n;
        scanf("%d",&n);
        for (int i = 0; i < n; n++)
        {
            int count = 0;
            char t[2][2];
            for (int i = 1; i <= 3; i++)
            {
                scanf("%c %c %c",&t[i][0],&t[i][1],&t[i][2]);
            }
            if (t[0][0] == '_' || t[0][1] == '_' || t[0][2] == '_' || t[1][0] == '_' || t[1][1] == '_' || t[1][2] == '_' || t[2][0] == '_' || t[2][1] == '_' || t[2][2] == '_')
            {
                for (int j = 0; j < 3; j++)
                {
                    if (t[j][0] == t[j][1] == t[j][2])
                    {
20)                        count++;
                    }
                }
                for (int q = 0; q < 3; q++)
                {
                    if (t[0][q] == t[1][q] == t[2][q])
                    {
27)                        count++;
                    }
                }
                if (t[0][0] == t[1][1] == t[2][2])
                {
32)                    count++;
                }
                else if (t[0][2] == t[1][1] == t[2][0])
                {
36)                    count++;
                }
                if (count == 0)
                {
                    printf("%d",2);
                }
                else if (count == 1)
                {
                    printf("%d",1);
                }
                else if (count > 1)
                {
                    printf("%d",3);
                }
            }
            
        }
    }

解决方法

下面的代码可能就是您要找的：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n,count=0; char s[20];
    printf("Enter total number of iterations (n): ");
    scanf("%s",s);    // Don't use %d. If you use %d and input a character,\
                          buffer issues arise.
    printf("\n");
    n = atoi(s);
    for (int i = 0; i < n; i++)
    {
        int count = 0;
        char t[3][3];
        for (int x = 0; x < 3; x++)
        {
            printf("Enter 3 chars with spaces between them: ");
            scanf(" %c %c %c",&t[x][0],&t[x][1],&t[x][2]);  // Space before\
                                                     the first %c makes scanf \
                                                     skip leading whitespace.
            printf("\n");
        }
    
        printf("n is %d\n\
t[0][0] = %c,t[0][1] = %c,t[0][2] = %c\n\
t[1][0] = %c,t[1][1] = %c,t[1][2] = %c\n\
t[2][0] = %c,t[2][1] = %c,t[2][2] = %c\n",n,\
\
t[0][0],t[0][1],t[0][2],\
t[1][0],t[1][1],t[1][2],\
t[2][0],t[2][1],t[2][2]);
        
        if (t[0][0] == '_' || t[0][1] == '_' || t[0][2] == '_' || \
            t[1][0] == '_' || t[1][1] == '_' || t[1][2] == '_' || \
            t[2][0] == '_' || t[2][1] == '_' || t[2][2] == '_')
        {
            for (int j = 0; j < 3; j++)
            {
                if ((t[j][0] == t[j][1]) && (t[j][1] == t[j][2]))
                {
                       count++;
                }
            }
            for (int q = 0; q < 3; q++)
            {
                if ((t[0][q] == t[1][q]) && (t[1][q] == t[2][q]))
                {
                       count++;
                }
            }
            if ((t[0][0] == t[1][1]) && (t[1][1] == t[2][2]))
            {
                   count++;
            }
            else if ((t[0][2] == t[1][1]) && (t[1][1] == t[2][0]))
            {
                   count++;
            }
            if (count == 0)
            {
                printf("%d",2);
            }
            else if (count == 1)
            {
                printf("%d",1);
            }
            else if (count > 1)
            {
                printf("%d",3);
            }
        }
        
    }
    return(0);
}

输出如下（count好像是> 1，所以打印了3）：

Enter total number of iterations (n): 1

Enter 3 chars with spaces between them: x x x

Enter 3 chars with spaces between them: o o o

Enter 3 chars with spaces between them: _ _ _

n is 1
t[0][0] = x,t[0][1] = x,t[0][2] = x
t[1][0] = o,t[1][1] = o,t[1][2] = o
t[2][0] = _,t[2][1] = _,t[2][2] = _
3