为什么在我的C代码中使用x-> y = z而不是xy = z？

Question

I'm following instructions to create code for a flexible array, and I am told to use x->y = z instead of xy = z. What would be the reason to use pointers instead of just assigning values normally?

This code includes a custom library flexarray.h which I'll put at the bottom

flexarray.c:

struct flexarrayrec{
   int capacity;
   int itemcount;
   int *items;
};

flexarray flexarray_new(){
   flexarray result = emalloc(sizeof *result);
   result->capacity = 2;
   result->itemcount = 0;
   result->items = emalloc(result->capacity * sizeof result->items[0]);
   return result;
}

flexarray.h:

#ifndef FLEXARRAY_H_
#define FLEXARRAY_H_

typedef struct flexarrayrec *flexarray;

extern void flexarray_append(flexarray f, int item);
extern void flexarray_free(flexarray f);
extern flexarray flexarray_new();
extern void flexarray_print(flexarray f);
extern void flexarray_sort(flexarray f);

#endif

Answer 1

x->y is just syntax sugar for this

(*x).y

Looks like flexarray is just typedef for *flexarrayrec , so when you write flexarray result , compiler will "transform" it to

flexarrayrec *result = emalloc(sizeof(flexarrayrec));

So result.capacity , for example, would have no meaning as pointer is just a number (memory address), "it doesn't have any field". (*result).capacity will work, but it's frustrating to write so many asterisks and parentheses, that's why we use -> operator :)

Answer 2

x->y is equivalent to (*x).y (it is syntactic sugar)

That is, if x is a pointer. If your structure (or object) is not a pointer, you can use xy (and not x->y)

In your case you must use dynamic allocation (pointers), since you want to use flexarray outside of the function flexarray_new().

Answer 3

Your flexarray is a pointer of struct flexarrayrec type. So-

typedef struct flexarrayrec *flexarray;

When you have structure pointer's you have to use -> arrow operator to access structure members!

But if you have declared

 typedef struct flexarrayrec flexarray; // Static declaration

Now you are allowed to use . dot operator to access its members!

If you want to use . means you can follow-

(*x).y = z instead of x->y = z;