Is (*x).y the same as x->y?

Question

Is operator -> allowed to use in C instead of .? Does its availability depend on compiler we are using? Is -> operator available in the last C standard or does it come from the C++ standard? How those two differ?

Solution 1

In C, c->m is equivalent to (*c).m. The parentheses are necessary since . has a higher precedence than *. Any respectable compiler will generate the same code.

In C++, unless -> or * is overloaded, the equivalence is as above.

Solution 2

There are 3 operators here, *, . and ->. This is important, because . and -> both have precedence 1, but * has precedence 2. Therefore *foo.bar is not the same as foo->bar and parenthesis are required, like in (*foo).bar.

All are original C operators and have been around forever.

Solution 3

In C, a->b and (*a).b are 100% equivalent, and the very reason for introduction of -> into C was precedence—so that you don't have to type the parentheses in (*a).

In C++, operator * and operator -> can be overridden independently, so you can no longer say that a->b and (*a).b are equivalent in all cases. They are, however, 100% equivalent when a is of a built-in pointer type.

Solution 4

Operator -> is standard in C. Both . and -> allow to access a struct field. You should use . on a struct variable and -> on a struct pointer variable.

struct foo {
    int x;
    int y;
}

struct foo f1;
f1.x = 1;
f1.y = 3;
struct foo *f2 = &f1;
printf("%d\n", f1.x);     // 1
printf("%d\n", f2->x);    // 1

The * operator is called dereference operator and returns the value at the pointer address. So (*f2).x is equivalent to f2->x.

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