CODEWITHSUNDEEP
c++pointers
return01
return01

Reputation: 13

Difference between pointer initializations

While I was reading C++ Primer, found this example that i don't understand

double dval;
double *pd = &dval; //ok:initializer is the address of a double
double *pd2 = pd; // ok:initializer is a pointer to double

Shouldn't *pd2 be a pointer to pointer since *pd is a pointer aswell? Could somebody explain what happens in the background with the memory addresses( what is assigned to pd2)?

Upvotes: 0

Views: 59

Answers (4)

E. Moffat
E. Moffat

Reputation: 3288

A pointer is a memory address. double *pd2 = pd; assigns one pointer to another pointer. No magic, just a simple assignment operation. Both pointers are pointing at the same memory address.

The value in pd2 is the same value as in pd. if you were to dereference either pd or pd2 you would get the value stored in dval.

Assigning &pd would require a pointer to a pointer since you are taking the address of a pointer.

Upvotes: 1

Nicolas Brauch
Nicolas Brauch

Reputation: 547

First you declare a double dval, then you declare a pointer, that points on that double (because its set by the doubles adress &dval). Then the second pointer is set equal to the first pointer. So finally, you have:

pd = pd2 = &dval

Which are two pointers, pointing to the same adress, not a pointer pointing to another pointer.

Upvotes: 0

Remy Lebeau
Remy Lebeau

Reputation: 596121

Shouldn't *pd2 be a pointer to pointer since *pd is a pointer aswell?

No. pd2 is being assigned the value that pd is currently holding. Thus pd2 ends up pointing at dval. You would only use a pointer-to-pointer if you want pd2 to point at pd itself, not what pd is pointing at, eg:

double dval;
double *pd = &dval;
double **pd2 = &pd;

Could somebody explain what happens in the background with the memory addresses( what is assigned to pd2)?

A pointer is just a number that happens to be a memory address. If you think of pointers as integers, the code you provided is not much different than the following behind the scenes:

double dval;
uintptr_t pd = address of dval;
uintptr_t pd2 = pd;

You are just assigning one pointer value (a number) as-is to another pointer (number) variable.

Upvotes: 1

g24l
g24l

Reputation: 3125

The value of pd2 is what is assigned to it, that is the address of dval, which was stored as value of pd and assigned to the former.

Upvotes: 1

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