Accessing value of array using pointers

Question

We know that we can dynamically create variables using pointers as for example:

int *p = new int(5);

We can access its value using * as

cout << *p;

But this is not the case of the array, consider the below code:

int size_array = 5;
int * p = new int[size_array];

if we need to access the value of the first element, we do the following:

cout << p[0];

but why we can't do the same as for a dynamic variable like above?, i.e. using *:

cout << *p[0];

p[index] is the same as *(p+index), so *p works just fine whether you use new int or new int[size_array], just note that in the second case it only accesses the 1st element in the array, unless p is updated to point at another element. Which you shouldn't do, use a separate pointer for that, if needed. — Remy Lebeau
– Remy Lebeau, Commented Dec 4, 2019 at 23:27

Acorn · Accepted Answer · 2019-12-05 07:36:18Z

5

p[0] is syntax sugar (for arrays) for:

*(p + 0)

Which is equivalent to:

*p

*p[0] does not work because it is equivalent to:

**p

In other words, de-referencing the pointer/array twice.

answered Dec 4, 2019 at 23:22

Acorn

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