Ttyjukl

In the C programming language, the syntax to access the member of a structure is

structure.member

However, a member of a structure referenced by a pointer is written as

pointer->member

There's really no need for two different operators. The compiler knows the type of the left-hand value; if it is a structure, the first meaning is evident. If it is a pointer, the second meaning is evident. Furthermore, . is far easier to type than ->. Not only does -> have more characters to type, on many keyboards one character is unshifted and the other character is shifted, requiring some finger acrobatics. Indeed, many languages based on C allow or use . in place of ->.

Why did C use two operators when one would have sufficed?

(My guess would be because C evolved from the typeless B language.)

I think there are two factors that led to standardization of the distinct operator "->" for accessing data members using a pointer.

1. You assume that the C compiler would recognize the type of the LHS as being a pointer. But programmers could, and often did, override the initial typing (variable declaration) by using a typecast.


2. In order to make the code more readable and less prone to unintended side-effects, it is useful to distinguish operations using pointers.

A very common feature of idiomatic C code is that a structure passed to a function as a pointer is modified within the function. Thus, the result is returned implicitly, by the side-effect of the structure variable in the calling function having been modified by the callee. This sort of approach violates modern sensibilities about loosely coupled code, but it was a simple and efficient means of dealing with complexity in C code. I would say the programmer was greatly assisted in maintaining the readability of such code by having distinct operations that made it clear whether some (possibly shared) memory pointer was the thing whose target was being modified.

In the embryonic form of C described in the 1974 C Reference Manual, there was no requirement that the left operand of . actually be a structure, nor that the left operand of -> actually be a pointer. The -> operator meant "interpret the value of the left operand as a pointer, add the offset associated with the indicated structure member name, and dereference the resulting pointer as an object of the appropriate type. The . operator effectively took the address of the left operand and then applied ->.

Thus, given:

struct q { int x, y; };

int a[2];

the expressions a[0].y and a[0]->y would be interpreted in a fashion equivalent to ((struct q*)&a[0])->y and ((struct q*)a[0])->y, respectively.

If the compiler had examined the type of the left operand to the . operator, it could have used that to select between the two behaviors for it. It was probably easier, however, to have two operators whose behaviors didn't depend upon the left operand's type.