Member access differences - c++

Question

can someone tell me what is the different between (*ptr).field and ptr->field?
I know it connect somehow to static and dynamic linking, but i dont know what is it.
can someone tell me the differnet and give me an example?
edit:
if i have this code:
Point p; //point is a class that derive from class shape
Shape *s=&p;
//there is a diffrence if i write:
(*s).print(); //print is virtual func
s->print(); // the answers will not be the same, why?
TNX!

Answer 1

This has nothing to do with static or dynamic linking.
See C++'s operator precedence. The . has a lower precedence than *, so there's actually quite a difference between *ptr.fld and ptr->fld. For example, the following code demonstrates:
#include <iostream>
struct foo {
int f;
};
int main() {
struct foo *p = new struct foo;
p->f = 42;
std::cout << p->f << std::endl;
std::cout << (*p).f << std::endl;
// The following will not compile
// std::cout << *p.f << std::endl;
}
As John Knoeller points out, ptr->fld is syntactic sugar for (*(ptr)).fld, but is not the same as *ptr.fld, which would actually evaluate to *(ptr.fld), probably not what you want.
You'd use ptr->fld when you have a pointer to a structure and want to access a field contained therein. (*(ptr)).fld means the same thing but is not as tidy. You'd use *strct.fld when you have a structure, not a pointer to a structure, that contains a field (fld) which is a pointer you want to dereference. The case of ptr->fld is shown above. The case of *strct.fld could be used with the following structure:
struct foo {
int *fld;
}
struct foo f;
f.fld = new int;
*f.fld = 42;

Answer 2

it has nothing to do with static or dynamic linking
both expressions will return the value of ptr.field
the ptr->field form is an abbreviated syntax for accessing a member directly from a pointer
UPDATE: it occurred to me that your original intent was not linking but binding
if this indeed was what you were aiming to then there is static binding and dynamic binding
which have some relation to the -> operator see here

Answer 3

Static linking is the result of the linker copying all library routines used in the program into the executable image. This may require more disk space and memory than dynamic linking, but is both faster and more portable, since it does not require the presence of the library on the system where it is run.
Dynamic linking is accomplished by placing the name of a sharable library in the executable image. Actual linking with the library routines does not occur until the image is run, when both the executable and the library are placed in memory. An advantage of dynamic linking is that multiple programs can share a single copy of the library.
But it is unrelated to pointer indirection you've mentioned -- in fact, those two expressions are identical.

Answer 4

I assume that by *ptr.field you meant (*ptr).field.
As far as only built-in operators are considered, there's no difference between the two. And no, it has nothing to do with and "static" or "dynamic" linking, whatever is implied by these terms.
The only potential difference between the two is that in ptr->field variant -> is an overloadable operator in C++, while in the (*ptr).field variant only * is overloadable, while . is not.
Also, some differences between these two methods of member access existed in very archaic versions of C language (CRM C), but I doubt anyone cares about those today.

Answer 5

This has nothing to do with linking.
ptr->fld
is merely shorthand for
(*ptr).fld
It's pure syntactic sugar ;)

Answer 6

As long as ptr is a pointer, the two are equivalent once properly parenthesized (as others have said). If ptr is an object, rather than a pointer, they might be different depending on definitions for operator* and operator-> (if any) from the object's class or ancestors.

Answer 7

The form -> is just shorthand for de-refrencing the pointer and accessing the member.
(*ptr).field;
// Equiv to
ptr->field;
One good reason for using -> is that when you are following a chain:
int x = (*(*(*(*ptr).field1).field2).field3).field4;
// Equiv to
int y = ptr->field1->field2->field3->field4;
The second one becomes much more readable.
As for the second part of your question.
I find it really easy to just nock up an example.
#include <iostream>
class Shape
{
public: virtual ~Shape() {}
virtual void Print() {std::cout << "Shape\n";}
};
class Point: public Shape
{
public: virtual void Print() {std::cout << "Point\n";}
};
int main ()
{
Point p;
Shape* s = &p;
s->Print();
(*s).Print();
}
> vi x.cpp
> g++ x.cpp
> ./a.exe
Point
Point
As you can see the result is the same in both situations.
When you call a method via a pointer or a reference the virtual call mechanism will be invoked. The star operator (AKA derefence operator) returns a reference to an object (it does not actually de-reference the object). So when it is used to call a method the virtual call mechanism will be invoked and the most derived version of the method called.

Answer 8

They are both the same.
*ptr.field dereferences the variable ptr then returns the value of member field.
The -> operator is shorthand notation for the above.