Question

This is probably a stupid question, but I don’t understand why this works:

 int** test = new int*[7];

 int x = 7;
 *(test+1) = &x;

 cout << (**(test+1));

test is a pointer to a pointer right? The second pointer points to the array, right?
In my understand I would need to dereference the “test” pointer first to get to the pointer that has the array.

(*test) // Now I have int*
*((*test) + 1) // to access the first element.

Where is my faulty thinking?

Answer 1

Is your misunderstanding that you think you have created a pointer to an array of 7 int? You haven’t. You actually have created an array of 7 pointers to int. So there is no “second pointer” here that would point to an array. There is just one pointer that points to the first of the 7 pointers (test).

And with *test you get that first pointer which you haven’t initialized yet, though. If you would add 1 to that, you would add 1 to some random address. But if you add 1 to test you get a pointer that points to the second pointer of the array. And dererencing that you get that second pointer, which you did initialize.

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What you describe would be achieved by a different syntax

typedef int array[7];
array* test = new int[1][7];

// Note: "test" is a pointer to an array of int. 
// There are already 7 integers! You cannot make it
// point to an int somehow. 
*(*test + 1) = 7;

int *p1 = *test
int i1 = *(p1 + 1); // i1 is 7, second element of the int[7]

delete[] test;

Without using the typedef, this looks like the following

int(*test)[7] = new int[1][7];

That is, you have created a one-element array, where the element-type of that is a 7-element array of int. new gives you a pointer back to that array. Note that the parenthesis is important: The * has less precedence than the [7], so otherwise this would be taken as an array of 7 pointer to integers.