Why do pointers use -> instead of .?

Question

Possible Duplicate:
Why does C have a distinction between -> and . ?

Lets say that I have this structure:

struct movies
{
    string title;
    int year;
} my_movie, *ptrMovie;

Now I access my_movie like this: my_movie.year = 1999;
Now to access a pointer I must do this: ptrMovie->year = 1999;

Why do pointers use the -> operator and normal data types use the . operator? Is there any reason they couldn't both use the . operator?

... because it would be confusing? Different operation -> different operator. — jldupont, Jan 07 '10 at 17:38
Duplicate: http://stackoverflow.com/questions/1813865/why-does-c-have-a-distinction-between-and — CB Bailey, Jan 07 '10 at 17:39
I suppose the 'duplicate' is a C question and not a C++ question but as much of the reason why C++ is grammar is like it is is because that's how C does it. Some answers in the other question also address C++ compatibility as well, though. — CB Bailey, Jan 07 '10 at 17:44
To be fair, if we look at C++ instead of C, another reason is that `operator->` needs to be distinct from operator. for implementing smart pointer classes (`operator->` must be overloadable to thunk through to the pointee, but `operator.` must be separate and non-overloadable to allow accessing methods on the smart pointer object itself). — jamesdlin, Jan 07 '10 at 19:18

score 9 · Accepted Answer · answered Jan 07 '10 at 17:39

9

The . operator accesses a member of a structure and can operate only on structure variables. If you want to do this to a pointer, you first need to dereference the pointer (using *) and then access the member (using .). Something like

(*ptrMovie).year = 1999

The -> operator is a shorthand for this.

answered Jan 07 '10 at 17:39

Noufal Ibrahim

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"The . operator accesses a member of a structure and can operate only on structure variables."??? really? how about class member variables? – Glen Jan 07 '10 at 17:43
I stand corrected. The OP mentioned structures and I confined my response to C (which doesn't have classes). – Noufal Ibrahim Jan 07 '10 at 17:46
@Noufal, yes, but the question is tagged C++ which has both structs and classes – Glen Jan 07 '10 at 17:50
1

Glen, there's no difference between a structure and a class outside their definition. – avakar Jan 07 '10 at 17:53
In C++ a structure is a type of class. `.` operates on all types of class including structures and unions. – CB Bailey Jan 07 '10 at 17:54
@Glen. Yup. My mistake. I'm not very familiar with C++ and that narrowed my answer as well. Your comment clarifies the point though. Thanks. – Noufal Ibrahim Jan 07 '10 at 17:57
@avakar, Charles, yes, I know that. The answer however states that the . operator can only be used on structs. This is obviously incorrect, so I'm just trying to clarify the answer for others. – Glen Jan 07 '10 at 17:59
@Glen: The `class` keyword creates a struct. However, some classes are not structs, they are unions, and the member access operators are usable with those too. – Ben Voigt Jun 07 '12 at 19:40

A. Levy · Answer 2 · 2010-01-07T18:19:29.807

4

The . operator is only valid for a struct or class. A pointer is not a struct or class, so you need to dereference your pointer to get the struct/class it is pointing to like this

(*ptrMovie).year

The member operator . has a higher precedence than the dereference operator *, so you need to enclose the dereferencing operation in parenthesis. Or you could do this

ptrMovie->year

Both are equivalent. The '->' operator is a shortcut for dereferencing your pointer and then accessing a struct member. It is less typing and a little nicer to use in my opinion. Apparently most people agree with me because that is the standard way to access struct members from a pointer to the struct in most code that I've seen. You especially appreciate the difference when you have to do multiple levels of indirection:

ptrToStruct->memberPtr->subMemberPtr->subsubPtr->subsubsubPtr->x

(*(*(*(*(ptrToStruct).memberPtr).subMemberPtr).subsubPtr).subsubsubPtr).x

Both of those statements are equivalent, but the first is easier to work with.

edited Jan 07 '10 at 18:19

answered Jan 07 '10 at 17:39

A. Levy

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". is only valid for a struct"??? really? how about classes? – Glen Jan 07 '10 at 17:42
Aside from the default protection (public for struct, private for class), class and struct are interchangeable in C++. – KeithB Jan 07 '10 at 17:55
@KeithB, yes, I know. However the answer implies otherwise. Maybe my comment wasn't particularly clear while trying to clarify that – Glen Jan 07 '10 at 18:00
Yeah...I should have said "struct or class" Even though they are functionally the same thing in C++. – A. Levy Jan 07 '10 at 18:10
How about `union`? In fact, it's valid for all classes (which are types introduced by the class-key `class`, `struct`, or `union`). – Ben Voigt Jun 07 '12 at 19:39

score 2 · Answer 3 · answered Jan 07 '10 at 17:39

If they both used . how could you differentiate between the pointer and the actual object? To me:

->

Reminds me of an arrow which points to something, so I find it great that -> is used.

Instead of typing (*myPointer). it is simplier to use myPointer->

Why do pointers use -> instead of .?

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