CODEWITHSUNDEEP
c++arrayspointers
sangmin park
sangmin park

Reputation: 567

Address of the Pointer pointing to the first element

In http://www.fredosaurus.com/notes-cpp/arrayptr/array-diagrams.html website, it states that

Pointers hold the memory address of other data and are represented by a black disk with an arrow pointing to the data it references.

For example, 

int a[5];  // Allocates memory for 5 ints.
. . .
a[0] = 1;
for (int i=1; i<5; i++) {
a[i] = a[i-1] * 2;
}

would result in. enter image description here

My question is how can I print the address of the pointer pointing to the array? I know that &a or &a[0] gives us the address of the first element. But how can I access the pointer pointing the array?

Upvotes: 2

Views: 2315

Answers (4)

eerorika
eerorika

Reputation: 238361

I know that &a or &a[0] gives us the address of the first element.

Technically. &a is in fact the address of the array itself. It just has the same value as address of the first element. But note that the type of the expression is different. &a is a int (*)[5] (pointer to an array of 5 integers), while &a[0] is a int* (pointer to an integer).

My question is how can I print the address of the pointer pointing to the array?

First step is to have a pointer. You don't have any pointer variables in your example. An array is not a pointer. The sentence "The actual array variable, a in this example, is a pointer to the memory for all of its elements." of the page is very misleading, as an array is not a pointer object. The drawn diagram is not correct for the program.

Here is a program that is correct according to the diagram:

int array[5];   // Allocates memory for 5 ints.
int* a = array; // a pointer to the first element of the array
a[0] = 1;
for (int i=1; i<5; i++) {
    a[i] = a[i-1] * 2;
}

Here we actually have a pointer. Now you can print the address of the pointer:

std::cout << &a;

Upvotes: 0

TheBeardedQuack
TheBeardedQuack

Reputation: 449

If the array is on the stack as it would be in your example code then the image below is a more accurate representation of how the memory would be laid out. I've also added an additional pointer to hopefully add some clarification.

int a[5];
a[0] = 1;
for (int i=1; i<5; i++)
    a[i] = a[i-1] * 2;

int* b = a;

It may be easier to think of the [] notation of defining an array as a little syntactic sugar. When calling functions and passing in a the function will be called with the address of a[0] being passed in.

When passing b to a function it's the value that will be passed in, which is the address of a.

Upvotes: 1

dbush
dbush

Reputation: 224072

An array and a pointer are not the same thing. What the memory really looks like is this:

   a
-------
|  1  |  a[0]
-------
|  2  |  a[1]
-------
|  4  |  a[2]
-------
|  8  |  a[3]
-------
| 16  |  a[4]
-------

So if you were to print a and &a, you would see that they print the same value.

Upvotes: 1

Matthieu Brucher
Matthieu Brucher

Reputation: 22023

You can have lots of pointers pointing to a[0].

Just as for a pointer to the array, create a pointer variable pointing to a[0] and then take a pointer to it. You will have one pointer to one pointer pointing to the array.

Upvotes: 0

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