CODEWITHSUNDEEP
c++arrayspointers
Mayuresh
Mayuresh

Reputation: 93

Array and pointers

int array[10] = {1,2,3,4,5,6,7,8};
int *ptr = array;
std::cout<<*(ptr+ 2); => O/P = 3

int (*arr_ptr)[10] = &array;
std::cout<<*(*arr_ptr+2); => O/P = 3

Whats the difference between these 2. I know what they mean, but if they give same result why do we have 2 different notations?

Upvotes: 3

Views: 188

Answers (4)

snapfractalpop
snapfractalpop

Reputation: 2134

They are not in fact the same to the compiler, and will result in different assembly language being generated.

Here is some good reference on the difference: http://cplusplus.com/forum/articles/10/

Upvotes: 0

iammilind
iammilind

Reputation: 70078

After editing your question, following are the differences:

(1) int *ptr = array;

array gets decayed to the pointer. ptr is ideally allowed to point to any int[] irrespective of its size

(2) int (*arr_ptr)[10] = &array;

arr_ptr is a pointer to an int[10]. It's very specific definition and you can never assign an array to arr_ptr which has size other than 10.

int array_2[20];
arr_ptr = &array_2; // error

Upvotes: 1

drdwilcox
drdwilcox

Reputation: 3951

Use the notation that you understand and that fits the problem domain. If your app has an array, then use array notation. If you have to push pointers around for some other reason, then the second version is appropriate.

Upvotes: 0

Cubbi
Cubbi

Reputation: 47468

The first case is the normal case: ptr is a pointer to the first element of the array array, and it is indexed (using *(ptr+2), which is a synonym for ptr[2]) to access the third element of that array.

The second case is contrived: arr_ptr is a pointer to the (enitre) array array. It is first dereferenced (with *arr_ptr) to yield the array itself, which is then used as an argument to binary +, which causes it to get implicitly converted to a (nameless) pointer to its first element, which is then indexed in the same way *(<nameless pointer>+2) and gives the same result.

Making those implicit conversions explicit, you could write

int *ptr = &array[0];
std::cout<<*(ptr+ 2);

int (*arr_ptr)[10] = &array;
std::cout<<*( &(*arr_ptr)[0] + 2 );

Upvotes: 3

Related Questions