How to use pointer-to-pointer's variables?

Question

I am trying this code:

int x[] = {1,5};
int y[] = {3,6};
int *w[] = {x,y};
int **z = w;


printf("%d", **z[1]);
cout<<**z[0];

But it doesn't work. My goal is to access to each element inside the x or y arrays. My question comes from this part of a really OpenCV code:

float h_ranges[] = {0, 180}; 
float s_ranges[] = {0, 256};
const float* ranges[] = {h_ranges, s_ranges};

And like to know how is it possible to read each element inside ranges[]?

EDIT: I tryed:

printf("%d", z[1][1]);
cout<<z[0][0];

And it worked, but it's interesting for me to know why

printf("%d", **z[1][1]);
cout<<**z[0][1];

Doesn't work, but the following code works

printf("%d", *w[1]);
cout<<*w[0];

What is the difference between w and z here?

EDIT2:

Why the result of

printf("%d", *w[1]);
cout<<*w[0];

Is same as the result of:

printf("%d", *z[1]);
cout<<*z[0];

Do you know what the type of z[1][1] is? If yes why do you think that adding two "stars" at the front is allowed? — UnholySheep
– UnholySheep, Commented Jun 16, 2018 at 10:05
@UnholySheep: But it is still confusing why *w[0] and *z[0] have same result? In fact I can not imagine truly the structure of z at this point! — lovecode
– lovecode, Commented Jun 16, 2018 at 11:00
**z[0] is dereferencing three times, that's why your cout does not work. — rpaulin56
– rpaulin56, Commented Jun 17, 2018 at 10:34
Let's recap: *x is another way of writing x[0], so **z is like z[0][0]. — rpaulin56
– rpaulin56, Commented Jun 17, 2018 at 10:37

Killzone Kid · Accepted Answer · 2018-06-16 20:08:24Z

3

Why z[1][1] works?

z is a pointer to pointer to int in w array. Dereferencing *z will give you pointer to the 1st int in {1,5} and dereferincing it once more **z will give you that int, which is 1. This is identical to doing z[0][0] or *(*(z + 0) + 0).

z[1][1] is the same as *(*(z + 1) + 1). Dereferencing *(z + 1) will give you the pointer to the first int in the second array {3,6}. Incrementing *(z + 1) + 1 will give you the pointer to the second int in the second array {3,6}, Dereferencing *(*(z + 1) + 1) will give you the actual second int of the second array {3,6}, which in your case is 6;

Why **z[1][1] doesn't work?

With z[1][1] will get you an int (see above), 6 in your example. So what happens when you try to dereference an int, let alone twice? An error!

Why *w[1] works?

w is array of pointers. With w[1] you get the second pointer from that array which points to the array of 2 ints, {3,6}. If you dereference it, you will get the 1st int of that array, which in your case is 3.

Why the result of *w[1] is same as the result of *z[1]?

Let's visualize it:

{1,5} {3,6}
 |     |
{w[0], w[1]}
 |     | 
 z     z+1

Array w is array of 2 pointers to int, which are w[0] and w[1]. w[1] is the second pointer to int in that array. The int it points to is 3. z is a pointer to pointer to int, or in this case pointer to w[0], which is pointer to 1. z[1] is the same as *(z+1) and it makes z to point to the next pointer before dereferencing result, which in this case is w[1], which points to 3. Dereferencing both w[1] and z[1] yields the same result, because ultimately they both point to the same int, which is 3.

edited Jun 16, 2018 at 20:08

answered Jun 16, 2018 at 12:00

Killzone Kid

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3 Comments

lovecode Over a year ago

Thank you very much! But may you explain more about *(*(z + 0) + 0). What is exactly meaning of z+0 or z+1 ? What is the z in this point that you +1 it? I think z is an address and you step to next address by +1.

lovecode Over a year ago

I also added a new question (EDIT 2) to my previous questions, that is vague for me!

Killzone Kid Over a year ago

@Hasani I have added a little visualisation. I'm afraid if you have any more questions after that you will have to post a new question.

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