Question: Here is my code:

```
std::vector< std::vector<std::shared_ptr<int>> > om(2, std::vector<std::shared_ptr<int>>(2));
om[0][0] = std::shared_ptr<int>(std::make_shared<int>(1));
om[0][1] = std::shared_ptr<int>(std::make_shared<int>(2));
om[1][0] = std::shared_ptr<int>(std::make_shared<int>(3));       //init values
om[1][1] = std::shared_ptr<int>(std::make_shared<int>(4));

std::vector<std::shared_ptr<int>>::iterator pd;                  //inner iterator
std::vector< std::vector<std::shared_ptr<int>> >::iterator px;   //outer iterator

for(px = om.begin(); px != om.end(); px++){                    
    for(pd = (*px).begin(); pd !=  (*px).end(); pd++){

    std::cout<< *pd[0] << std::endl;
    }

}

```

Output:

> 
> 1 2 3 4
> 
> 
> 

I am making a 2d vector using `shared_ptr`, and my goal is to print the values

My questions:

From the other examples that i have seen, you only need to dereference the iterator to get the value, but in my case if i will use `std::cout<< *pd << std::endl;` it will print raw addresses which isn't what i want.

But if i will use `std::cout<< *pd[0] << std::endl;` then it will print the right values.

i visualize it like this : 

```
[ [1][2], [3][4] ]

```

wherein px will iterate 2 blocks and pd will have to iterate 4 blocks in total. 

Why do i have to put `[0]` in order to print? or is that undefined?

Answer: 
Since `pd` is of type `std::vector<std::shared_ptr<int>>::iterator`, `*pd` is of type `std::shared_ptr<int>`. You must dereference once more to get the integer value.

`*pd[0]` is equivalent to `*(*(pd+0))`, which is in turn equivalent to `**pd`.
> 
> Why do i have to put [0] in order to print? or is that undefined?
> 
> 
> 

`std::vector::iterator` is a random access iterator. That provides `operator[]` with the same semantics as for a raw pointer. So, `p[0]` has the effect of de-referencing the *iterator*, giving you a `shared_ptr` lvalue reference. You then de-reference that with `*`, giving you the `int` it "points" to.

You could have done this too, which might have been easier to follow:

```
std::cout<< **pd << std::endl;

```