C中的指针数组和结构

Question

I am nowhere lose to even guessing how come we get the given output . Can I please get some explanation? Also, any short and quick resources to try similar questions?

void main()
{
    struct a
    {
        char ch[10];
        char *str;
    };

    struct a s1={"Hyderabad","Bangalore"};
    printf("\n%c%c",s1.ch[0],*s1.str);
    printf("\n%s%s",s1.ch,s1.str);
    getch();
}

Ans: HB, HyderabadBangalore

Answer 1

struct a s1={"Hyderabad","Bangalore"}; assigns "Hyderabad" to ch and "Bangalore" to str .

printf("\\n%c%c",s1.ch[0],*s1.str); prints the first character of the strings. Since ch is an array, ch[0] represents its first character. Since str is a character pointer, it points to the first character of a string here. So, *s1.str will have value 'B'

printf("\\n%s%s",s1.ch,s1.str); simply prints all characters of both strings. Fundamentally, ch is equal to &ch[0] , the address of the first character in the array. And, str is pointer variable which holds the address of first character of the string literal "Bangalore".

Answer 2

%c is only for single character. Example H.
$s is for string. Example Hyderabad.

s1.ch[0] points to first character of String ch ->H
*s1.str is a pointer. It points to the value stored at address of str. It will be ->B
Hence you get HB

\\n means a new line (as in java).

Answer 3

Here is your output according to requested question

void main()
    {
        struct a
        {
            char ch[10];
            char *str;
        };

        struct a s1={"Hyderabad","Bangalore"};
        printf("%c%c",s1.ch[0],*s1.Str[0]);
        printf("\t%s%s",s1.ch,s1.str);
        getch();
    }