Using dot operator vs arrow operator with string pointer in structs

Question

Q1: I'm trying to understand why in this code I'm supposed to use a dot operator when calling the string pointer rather than an arrow which is used for pointers.

Q2: If I was trying to initialize age would I use the dot operator?

typedef struct Person 
{
  char *name;
  int age;

} Person;

Person* deletePerson(Person *person, int totalPeople) 
{
     int i;

     for(i=0; i<totalPeople; i++) 
     {
     free(person[i].name); 
     free(person); 
     }

     return NULL;
}

The only thing one needs to know about it is that `a->b` is the same as `(*a).b` and the same as `a[0].b`. — Eugene Sh., Sep 06 '19 at 17:57
`person` is, depending on how you think about it, either an array of `struct Person`, or a pointer to `struct Person`. So `person[0]` is a `struct Person`. Since it's a struct (*not* a pointer-to-strut), you use `.` on it. — Steve Summit, Sep 06 '19 at 17:57
There's something else wrong with this code: `free(person)` is almost certainly wrong, and especially so given that it's in the loop. — Steve Summit, Sep 06 '19 at 17:59
It depends on what's on the LHS of the operand. You use `pointer->age` or `structure.age`. `person[i]` is a structure; use `person[i].age`. `person` alone is a pointer; use `person->age`. — Jonathan Leffler, Sep 06 '19 at 18:00
In answer to your other question, you would access `age` the same way you access `name`: perhaps as `person[i].age`. And, in fact, to confirm what's going on, you might want to print out `person[i].name` and `person[i].age` as you free them. — Steve Summit, Sep 06 '19 at 18:01
The name of the variable is semantically wrong. It should be `persons` or `people`... — Antti Haapala -- Слава Україні, Sep 06 '19 at 18:05
@AnttiHaapala: But `person[i]` reads better than `persons[i]`. — Scott Hunter, Sep 06 '19 at 18:06
@ScottHunter that is true... but there is no justification for the function name. Nor for the return value — Antti Haapala -- Слава Україні, Sep 06 '19 at 18:08
Why you need the loop? If you already have the person pointer you can just `free(person->name); free(person);` — André Sanson, Sep 06 '19 at 18:18
Related/Duplicate [Why does the arrow (->) operator in C exist?](https://stackoverflow.com/questions/13366083/why-does-the-arrow-operator-in-c-exist) — David C. Rankin, Sep 06 '19 at 18:23
Unless I'm missing something here `person[i].name`will throw error — André Sanson, Sep 06 '19 at 18:31

score 4 · Answer 1 · answered Sep 06 '19 at 18:07

You need to use the arrow -> operator when you have a pointer to a struct (or union) on the left, and the dot . operator when you have a struct on the left. It doesn't depend on what's on the right.

If person was a pointer to a single Person, to access its field, you'd use person->name and person->age.

But here person is evidently a pointer to the first element of an array of Person. (A pointer to the first element is how arrays are passed around in C.) To access an individual element of the array, you use the subscript operator (square brackets […]): person[i]. person[i] is the element with index i, not a pointer to the element with index i. Since person[i] is a struct, you use the dot operator to access its fields.

In addition, as others have already remarked, free(person); inside the loop doesn't make sense. You'd free the whole array as soon as you've processed the first element. Call free(person) after the loop, once you've finished cleaning up all the elements of the array (assuming that person was allocated with malloc).

Scott Hunter · Answer 2 · 2019-09-06T18:04:10.583

person->name would be synonymous with person[0].name; since you want to refer to the name of an arbitrary person, you couldn't use -> this way.

(I suppose you would use (person+i)->name, but your original code is clearer, IMHO.)

As for initializing age: what else would you use other than .?

P.S. freeing person inside the loop means a) you free it multiple times, and b) person[i] will be a problem after the first iteration.

score 0 · Answer 3 · answered Sep 06 '19 at 18:04

For starters the function is invalid and has a bug.

I think you mean

Person* deletePerson(Person *person, int totalPeople) 
{
     for ( int i=0; i < totalPeople; i++ ) 
     {
         free( person[i].name ); 
         free( person[i] ); 
     }

     free( person );

     return NULL;
}

This statement

free(person[i].name);

can be equivalently rewriten like

free( ( person + i )->name );

or even like

free( ( *( person + i ) ).name );

However the first statement is more readable.

According to the C Standard (6.5.2.1 Array subscripting)

2 A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).

Tal Avraham · Answer 4 · 2019-09-06T18:46:08.880

In C, the -> operator is a cosmetic substitute for (*var).field, and the [ ] is a substitute for *(ptr + n).

In your code, the line:

free(person[i].name);

Is equivalent to:

free((*(person + i)).name)

If you use -> instead of the dot operator that would be wrong. I can see why you are confused, to make it look simpler, the function parameter “Person *person” is basically an array of struct Person, equivalent to “Person person[]”. If you were to access only the first Person in the array you could have used person->name which is the same as person[0].name

Using dot operator vs arrow operator with string pointer in structs

4 Answers4