Why can I access a structure without the pointer?

Question

When I declare a pointer to struct I can access members without entering an address struct, why is it possible? For example:

struct exemplo{
    int a ; 
};

int main()
{
    struct exemplo *p;

    p->a = 10 ;    
}

Answer 1

The operator -> is the pointer access of a member of a structure. It is syntactic sugar for (*p).a, in the shown example.

The code has flaws. It even works, but may not do what you expect. Missing declare the type for the structure (the code gets cleaner) and allocate memory to accommodate it. The correct code would be:

#include <stdio.h>
#include <stdlib.h>
typedef struct {
   int a;
} Exemplo;

int main(void) {
    Exemplo *p = malloc(sizeof(Exemplo));
    p->a = 10;    
    printf("valor = %d", p->a);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

It doesn’t get better this way?

If you do not want to declare the variable for the structure, you can also do it, but then you have to be explicit for all use of it, see how. Credits to the Carlosfigueira for having commented on this.

If you really want to put the structure on stack (not use malloc(), so don’t use pointer:

#include <stdio.h>
typedef struct {
   int a;
} Exemplo;

int main(void) {
    Exemplo p = { 10 };
    printf("valor = %d", p.a);
    Exemplo p2 = { .a = 10 }; //apenas uma variação
    printf("valor = %d\n", p2.a);
    Exemplo p3;
    p3.a = 10; //outra variação
    printf("valor = %d\n", p3.a);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

It is possible to initialize with pointer as well.