Question

免责声明 - 这是一些功课，所以我不想要答案，但没有人去寻求指导（除非我要等一个星期，我通过函授学习。）

无论如何，我正在写一个小扫雷功能。用户输入网格的大小和地雷的数量。然后我打印出电路板和电路板。结果。例如，在有3个地雷的3 * 3电路板上，我会打印：

...
*..
..*

110
*11
10*

*显然是地雷。除了结果之外，一切都很顺利。我的比较似乎不起作用。我正在使用==符号比较字符，这看起来是完全合法的方法，但结果对我没有意义。

我不想要答案，只是朝着正确的方向前进。可能是因为我在我设置的原始数组的边界之外比较我的数组。如果是这种情况，那么我不知道如何解决这个问题。

我的代码如下。谢谢！

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

void number1(int *stats);
void number2(int *stats);

int main(void)
{
   int n,m,i,j,k,x, rand1, rand2, counter, numMines;
   char choice;

do

{
   printf("------------\n");
   printf("lines (m) = : ");
   scanf("%d", &m);
   printf("\ncolumns (n) = : ");
   scanf("%d", &n);
   printf("\nMines = :");
   scanf("%d", &numMines);

   // check to ensure that numMines is not > n * m
   if (numMines > m * n )
   {
      printf("Error. You cann't have more mines than available spaces ");
      break;
   }

   counter = 0;
   //create minefield array

   char mineGrid[n][m];

    //set all fields in minesweeper to safe (.)

   for (j = 0; j < n; j++)
   {
      for (i = 0; i < m; i++)
      {
         mineGrid[j][i] = '.';
      }
    }
    // Need to randomally make some mines for the array
    //bounds of the random numbers is defined by the values of n and m

   for (i = 0; i < numMines; i++)
   {
      //generate random number
      rand1 = ((double)rand() / ((double)RAND_MAX + 1) * n);
      rand2 = ((double)rand() / ((double)RAND_MAX + 1) * m);
      //set as a mine:
      mineGrid[rand1][rand2] = '*';

      //Note : at the moment, mines can overlap. need to come back to fix this up.
    }
   //print out the msweeper grid
   x = 0;
   for (j = 0; j < n; j++)
   {
      printf("\n");
      x++;
      for (i = 0; i < m; i++)
      {
         printf("%c", mineGrid[j][i]);
      }
         printf("     ");
    }

   // here is where I will print the results:
   printf("\n");
   for (j = 0; j < n; j++)
   {
      for (k = 0; k < m; k++)
      {

      if (mineGrid[x+0][k+1]=='*')
         counter++;
      if (mineGrid[x+0][k-1]=='*')
         counter++;
      if (mineGrid[x+1][k+0]=='*')
         counter++;
      if (mineGrid[x+1][k-1]=='*')
         counter++;
      if(mineGrid[x+1][k+1]=='*')
         counter++;
      if (mineGrid[x-1][k+0]=='*')
         counter++;
      if (mineGrid[x-1][k-1]=='*')
         counter++;
      if(mineGrid[x-1][k+1]=='*')
         counter++;

      printf("%d", counter);
      counter = 0;
      }

   printf("\n");

   }

   printf("\n\nTry again? (y/n:) ");
   scanf("\n%s", &choice);

}while(choice == 'y' || choice == 'Y');
}

Answer 1

正如0A0D所指出的那样，您使用x来表示您要使用j，但您也可以自由地从坐标中添加和减去1，而不检查它们是否仍在场。那是在惹麻烦。

C - 将字符数组与char进行比较

1 个答案: