CODEWITHSUNDEEP
cpointers
Programm-ist
Programm-ist

Reputation: 109

Simple C double pointer

I am learning C pointers and quite confused. I tried to search online but couldn't find any clear explanation. This is what I am trying to understand:

int x = 8;
int *ptr = &x;
int **dptr = ptr;

What does **dptr point to, x or *ptr? When I tried to print the content of dptr I found that it contains the address of x instead of *ptr but I am not sure why?

Edited

int x = 8;
int *p = &x;
int **ptr = &p;
int **dptr = ptr;

Upvotes: 0

Views: 167

Answers (3)

John Bode
John Bode

Reputation: 123448

Given the declarations

int x = 8;
int *p = &x;
int **ptr = &p;
int **dptr = ptr;

Then the following are true:

**dptr == **ptr == *p ==  x == 8  (int)
 *dptr ==  *ptr ==  p == &x       (int *)
  dptr ==   ptr == &p             (int **)

Upvotes: 0

Arun A S
Arun A S

Reputation: 7006

int x = 8;
int *p = &x;
int **ptr = &p;
int **dptr = ptr;

First, the value 8 is stored in x.

Next, with the statement

int *p = &x;

p will point to x ( that is , p now stores the address of x )

Next

int **ptr = &p;

Now ptr will store the address of p, which in turn stores the address of x

int **dptr = ptr;

dptr stores the address of ptr which in turn stores the address of p which points to x.

So, Now, **dptr will have the value of the integer x.

Upvotes: 0

ouah
ouah

Reputation: 145829

First off, I assume you meant: int **dptr = &ptr; as int **dptr = ptr; is invalid.

int x = 8;
int *ptr = &x;
int **dptr = &ptr;

dptr points to ptr object.

*dptr points to x object.

**dptr is not a pointer but the int object x.

EDIT: question was edited, ptr was an int * but seems to be an int ** now...

Upvotes: 2

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