Pointer of array, what is the difference between (*ptr)[] and *ptr[]

Question

I met a weird thing when I tried to understand the pointer tp this array

#include <iostream>

int main() {
    using namespace std;

    short tell[3]{1, 2, 3};

    short (*pas)[3] = &tell;

    cout << (*pas)[2] << endl;
    cout << *pas[2] << endl;

    cout << endl;

    return 0;
}

I got two different values for the two outputs.

The first one is correct, which is 3.

However, for the second one, it seems it returns a random number which is different every time.

What is the difference between these two?

Answer 1

You declared a pointer to a whole array

short (*pas)[3] = &tell;

So to get the pointed object (array) you have to dereference the pointer like

*pas

Now this expression yields a lvalue reference to the array tell. So you may apply the subscript operator to access elements of the array

cout << (*pas)[2] << endl;

The postfix subscript operator has a higher priority than the unary operator *.

that is this expression

*pas[2]

is equivalent to the expression

*( pas[2] )

It means you are trying to access an object (array) beyond the allocated array that results in undefined behavior.

If you had a two=dimensional array like for example

short tell[3][3] =
{ { 1, 2, 3 },
  { 4, 5, 6 },
  { 7, 8, 9 }
};

And initialized the pointer pas like

short (*pas)[3] = tell;

the the expression pass[2] would yield the third element of the array that is { 7, 8, 9 } In this case you may apply one more subscript operator to access an element of the array like for example

pass[2][2]

that contains the value 9.

The above expression also can be rewritten without the subscript operator like

*( *( pass + 2 ) + 2 )