What How to Use Instead of the Arrow Operator, '-≫'

What can I use instead of the arrow operator, `->`?

The following two expressions are equivalent:

a->b

(*a).b

(subject to operator overloading, as Konrad mentions, but that's unusual).

Arrow operator (->) usage in C

foo->bar is equivalent to (*foo).bar, i.e. it gets the member called bar from the struct that foo points to.

Why do we use the dot operator (.) instead of the arrow operator (->) in arrays of structs as function parameters?

The array index operator [] contains an implicit pointer dereference. So if array has either array-of-struct or pointer-to-struct type, then array[0] has struct type and not pointer type.

Why does the arrow (->) operator in C exist?

I'll interpret your question as two questions: 1) why -> even exists, and 2) why . does not automatically dereference the pointer. Answers to both questions have historical roots.

Why does -> even exist?

In one of the very first versions of C language (which I will refer as CRM for "C Reference Manual", which came with 6th Edition Unix in May 1975), operator -> had very exclusive meaning, not synonymous with * and . combination

The C language described by CRM was very different from the modern C in many respects. In CRM struct members implemented the global concept of byte offset, which could be added to any address value with no type restrictions. I.e. all names of all struct members had independent global meaning (and, therefore, had to be unique). For example you could declare

struct S {
  int a;
  int b;
};

and name a would stand for offset 0, while name b would stand for offset 2 (assuming int type of size 2 and no padding). The language required all members of all structs in the translation unit either have unique names or stand for the same offset value. E.g. in the same translation unit you could additionally declare

struct X {
  int a;
  int x;
};

and that would be OK, since the name a would consistently stand for offset 0. But this additional declaration

struct Y {
  int b;
  int a;
};

would be formally invalid, since it attempted to "redefine" a as offset 2 and b as offset 0.

And this is where the -> operator comes in. Since every struct member name had its own self-sufficient global meaning, the language supported expressions like these

int i = 5;
i->b = 42;  /* Write 42 into `int` at address 7 */
100->a = 0; /* Write 0 into `int` at address 100 */

The first assignment was interpreted by the compiler as "take address 5, add offset 2 to it and assign 42 to the int value at the resultant address". I.e. the above would assign 42 to int value at address 7. Note that this use of -> did not care about the type of the expression on the left-hand side. The left hand side was interpreted as an rvalue numerical address (be it a pointer or an integer).

This sort of trickery was not possible with * and . combination. You could not do

(*i).b = 42;

since *i is already an invalid expression. The * operator, since it is separate from ., imposes more strict type requirements on its operand. To provide a capability to work around this limitation CRM introduced the -> operator, which is independent from the type of the left-hand operand.

As Keith noted in the comments, this difference between -> and *+. combination is what CRM is referring to as "relaxation of the requirement" in 7.1.8: Except for the relaxation of the requirement that E1 be of pointer type, the expression E1−>MOS is exactly equivalent to (*E1).MOS

Later, in K&R C many features originally described in CRM were significantly reworked. The idea of "struct member as global offset identifier" was completely removed. And the functionality of -> operator became fully identical to the functionality of * and . combination.

Why can't . dereference the pointer automatically?

Again, in CRM version of the language the left operand of the . operator was required to be an lvalue. That was the only requirement imposed on that operand (and that's what made it different from ->, as explained above). Note that CRM did not require the left operand of . to have a struct type. It just required it to be an lvalue, any lvalue. This means that in CRM version of C you could write code like this

struct S { int a, b; };
struct T { float x, y, z; };

struct T c;
c.b = 55;

In this case the compiler would write 55 into an int value positioned at byte-offset 2 in the continuous memory block known as c, even though type struct T had no field named b. The compiler would not care about the actual type of c at all. All it cared about is that c was an lvalue: some sort of writable memory block.

Now note that if you did this

S *s;
...
s.b = 42;

the code would be considered valid (since s is also an lvalue) and the compiler would simply attempt to write data into the pointer s itself, at byte-offset 2. Needless to say, things like this could easily result in memory overrun, but the language did not concern itself with such matters.

I.e. in that version of the language your proposed idea about overloading operator . for pointer types would not work: operator . already had very specific meaning when used with pointers (with lvalue pointers or with any lvalues at all). It was very weird functionality, no doubt. But it was there at the time.

Of course, this weird functionality is not a very strong reason against introducing overloaded . operator for pointers (as you suggested) in the reworked version of C - K&R C. But it hasn't been done. Maybe at that time there was some legacy code written in CRM version of C that had to be supported.

(The URL for the 1975 C Reference Manual may not be stable. Another copy, possibly with some subtle differences, is here.)

When to use arrow and when dot?

The array index operator [] has a dereference built into it.

g[i] is exactly the same as *(g + i). So g[i] refers to a DOCUMENT, not a DOCUMENT * and thus you use the member access operator . instead of the pointer-to-member operator ->.

Why do certain C programs use an arrow operator to point to parts of a structure instead of directly referencing?

In C/C++, the -> operator is used to access the props and functions of an object that a pointer is pointing at (ie. myClass->propOne). Keeping in mind that a pointer is just a reference to memory, you can see that it would not have propOne since it is just a memory location. The -> operator says that you want to access propOne of the object located in the memory that myClass is pointing at.

You could also achieve this by dereferencing the pointer with the * operator to access the object, and then use the . operator to access the prop (ie. (*myClass).propOne).

They are both valid, but the -> is a bit cleaner, and can lead to fewer coding errors.

Addressing the why question, there are a lot of reasons to use pointers. Sometimes you want to pass around a value or a structure but you do not want to copy it. Sometimes you may want more than one thing to be able to access the same structure easily. Sometimes it is just easier and cleaner.

C++ arrow operator equivelence

You've got the associativity rules wrong. a->b->c is (a->b)->c, not a->(b->c), so it becomes (*(a->b)).c (and then (*((*a).b)).c).

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