Pointers and structures

Question

I'm trying to wrap my head around pointers, references and addresses but every time I think I got it something unexpected pops up.

Why don't we need to dereference the structure to set a value in this example?

// pointer_tet.cpp
 #include <iostream>
struct example
{
 char name[20];
 int number;
};
int main()
{
 using namespace std;
 example anExample = {"Test", 5};
 example * pt = &anExample;
 pt->number = 6;
 cout << pt->number << endl;
 int anotherExample = 5;
 int * pd = &anotherExample;
 *pd = 6;
 cout << *pd << endl;
 return 0;
}

Thanks!

Edit: Thank you for your answers! What confused me was not being able to set *pt.number = 6.

• Seth Carnegie: I don't see anywhere you don't dereference to set a value.
• AndiDog: Umm, you are dereferencing pt.
• tletnes: where do you think you do not derefrence?

Answer 1

You are dereferencing pt. You are doing:

pt->number = 6;

This is equivalent to:

(*pt).number = 6;

The -> operator provides a convenient way to access members through a pointer.

• Q-bertsuit: Thank you! I read this the other day, but it already slipped my mind. Thanks! Edit: I have to wait 10 min to set answer
• Yakk - Adam Nevraumont: The trick here is that *pt.number is *(pt.number) -- way back in the day, the precedence of * was set awkwardly, and -> was added to make up for having to type (*pt).number all the time.

Answer 2

You can do

anExample.number = 6;

OR

(*pt).number = 6;

Read cplusplus.com pointer tutorial might help.