Question

Below is my code snippet

struct encode
{

   char code[MAX];

}a[10];

int main()
{ 
char x[]={'3','0','2','5','9','3','1'};

  for(i=0;i<1;i++)
  {

      printf("%c",x[i]);

//This will printout like 3025931 now I want this to be stored in structure.         
  }

strcpy(a[0].code,x); 
// or 
a[0].code=x;//neither works

display();

}

void display()
{
printf("%c",a[0].code);
}

I want the output to be like:3025931.

Which I am not getting due to incompatible assign type. Please tell me where am i going wrong.

Answer 1

I see two problems here. The first is that the source of the strcpy is a where it probably should be x.

The second is that x is not null-terminated. Strings in C are null-terminated character arrays.

I would change the two lines:

char x[] = {'3','0','2','5','9','3','1'};
strcpy(a[0].code, a);

to:

char x[] = {'3','0','2','5','9','3','1', '\0'};
strcpy(a[0].code, x);

Here’s a complete program that gives you what you want (it actually prints out the number twice, once in your inner loop character by character and once with the printf so that you can see they’re the same):

#include <stdio.h>
#include <string.h>
#define MAX 100
struct encode {
    char code[MAX];
} a[10];

int main() {
    int i, j;
    char x[] = {'3','0','2','5','9','3','1','\0'};

    for(i = 0; i < 1; i++) {
        for(j = 0; j < 7; j++) {
            printf("%c", x[j]);
        }
        printf("\n");

        strcpy(a[0].code, x);
    }
    printf("%s\n",a[0].code);
    return 0;
}

Update based on comment:

I am sorry. I am new to C. My apologies for not pasting the code snippet correctly in the beginning: “printf(“%c”,a[0].code);” doesn’t display “3025931”.

No, it won’t. That’s because a[0].code is a character array (string in this case) and you should be using "%s", not "%c". Changing the format specifier in the printf should fix that particular issue.