Question

代码是做康威的生活游戏，读取文件中的点并且可以正确打印，一旦做了规则，它只会打印我的板。如果我删除规则部分，它会给我正确数量的nb和位置。

#include <stdio.h>
#include <stdlib.h>
int l, w, i, n, a, x, y;
char b[80][80], ip[10];
int main(int argc, char *argv[])
{  

   n=argc;/*check the number of elements in argv*/
   if (n!=5){
      fprintf(stderr, "ENTERED ELEMENTS IS WRONG\n");
      exit (0);
   }

   w=atoi(argv[2]);/*read from a file*/
   l=atoi(argv[3]);

   FILE* f;
   f = fopen(argv[1],"r");
   if (f == NULL){/*print ERROR when file the is null*/
      fprintf(stderr, "ERROR READING FILE\n");
      exit (0);
   }

   for (x = 0; x < l; x++){/*all points are 0*/
       for(y = 0; y < w; y++){
          b[x][y] = 0;
        }
   }
   fscanf(f, "%d", &a);/*get the number of points*/
   for (i=0; i<a; i++){
       fscanf(f, "%d %d", &x, &y);
       b[x][y] = 1;/*the points that mentioned are 1*/
   }

   printf("*");/*To print the first line of the board*/
   for (i=0; i< w; i++){
       printf("-");
     }
   printf("*\n");

   for (x=0; x<l; x++){
        printf("|");
        for (y=0; y<w; y++){
            if (b[x][y]==1){
            printf("X");/*frint all the ponints, 0 is ' ', 1 is 'X'*/
            }
            else {
            printf(" ");
            }
        }
        printf("|\n");
     }

   printf("*");/*To print the last line of the board*/
   for (i=0; i< w; i++){
       printf("-");
     }
   printf("*\n\n");
/*the end of printing the first graph*/

   fgets(ip,10,stdin);
   while (ip!= NULL){
      generate();
      fgets(ip,10,stdin);
   }
}

void generate(void){
   int nb;
/*4 rules*/
   for (x = 1; x < l-1; x++){
       for(y = 1; y < w-1; y++){
         nb = b[x-1][y-1]
             +b[x-1][y]
             +b[x-1][y+1]
             +b[x][y-1]
             +b[x][y+1]
             +b[x+1][y-1]
             +b[x+1][y]
             +b[x+1][y+1];
         if (b[x][y]==1){
            if (nb<2||nb>3){
               b[x][y]=0;
            }else {
                b[x][y]=1;
            }
         }
         if (b[x][y]==0){
            if (nb==3){
               b[x][y]=1;
            }
         }
       }
   }

/*To print the board  same as printing the graph in main function*/
   printf("*");
   for (i=0; i< w; i++){
       printf("-");
     }
   printf("*\n");

   for (x=0; x<l; x++){
        printf("|");
        for (y=0; y<w; y++){
            if (b[x][y]==1){
            printf("X");
            }
            else {
            printf(" ");
            }
        }
        printf("|\n");
     }

   printf("*");
   for (i=0; i< w; i++){
       printf("-");
     }
   printf("*\n\n");
}

Answer 1

你需要两个板：当前的和未来的（下一个）板：char b[80][80], next[80][80]。评估规则时，请从当前板（b）获取数据，但将结果记录到未来的板（next[..][..]=...）中。在之后，您遍历所有单元格，将未来电路板的内容复制到当前电路板中bcopy()或嵌套循环。

Conway的生命游戏中的生命逻辑错误

1 个答案: