CODEWITHSUNDEEP
c++
Q-bertsuit
Q-bertsuit

Reputation: 3437

Pointers and structures

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.

Upvotes: 1

Views: 103

Answers (2)

SDEZero
SDEZero

Reputation: 363

You can do 

anExample.number = 6;

OR

(*pt).number = 6;

Read cplusplus.com pointer tutorial might help.

Upvotes: 0

Joseph Mansfield
Joseph Mansfield

Reputation: 110658

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.

Upvotes: 8

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