Why this char Array(C language) prints more than its capacity?

Question

I have this int count[1]; this array gives me 3 Rab output! but it has only 2 capacity and i wondered why?!

struct MEMBERS
{
    int code;
    char Fname[40];
    char Lname[40];
    int Pnum[20];
    float bedeh;
    float credit;
    int count[1];
    struct meroDate issued;
    struct meroDate duedate;
};
struct MEMBERS b;

int main()

{
  int i = 0, line = 5;
  char w[100];
  int str[100];
                FILE *myfile;
                myfile = fopen("Members.txt","r");
   if (myfile== NULL)
      {
         printf("can not open file \n");
         return 1;
      }

      while(line--){
                fgets(str,2,myfile);
                strncpy(b.count, str, 1);
              i++;
             printf("%s",b.count);
                   }

         fclose(myfile);

         return 0;

 }

my Members.txt contains:

3
Rebaz salimi 3840221821 09188888888
4
95120486525
95120482642

Answer 1

count is an array of integers which can hold only one element. You pass it to functions which required char * as arguments thus, causing undefined behaviour.

printf("%s",b.count);      // %s expects char * not int[]

Also here-

strncpy(b.count, str, 1);

You need to manually copy using loop or use sscanf() depending on your data.

Edit :-

  fgets(str,2,myfile);
        if(sscanf(str, "%d", &b.count[0])==1){
               printf("%d\n", b.count[0]);
               i++;
    }       

read from file and store in b.count[0] , check return of sscanf if successful then print it's value.

Also,

fgets(str,2,myfile);
      ^^^ str is int[] but fgets requires argument as char *. 

So str should be of type char * .