Nested structs and the usage of this

[arczi w]

Suppose I have the following code:

struct N
{
int a;
struct NN
{
int b;
};
};

Is there a way to write a function in the inner struct that would return a pointer to the outer struct?

[mbozzi]

There are only minor differences between classes that are nested and those that aren't.

struct N
{
  int a;
  struct NN { N* get_n() { return new N; } }; // e.g.
};

int main() 
{
    N::NN nn; 
    delete nn.get_n();
}

[arczi w]

You seem to create a new pointer to N whereas what I had in mind is returning a pointer to an outer structure for which memory was already allocated, something like this but it would need to point to N instead of NN.

[mbozzi]

The point is that N::NN has no association with N. These classes are completely independent - you might have written

struct N { int a; };
struct NN { int b; };

to the same effect.

If N contained a member variable of type N::NN instead, we could get closer:

#include <cassert>

struct N 
{
  struct NN 
  { 
    explicit NN(N* outer)
      : outer_(outer) 
    {}

    N* outer_;
  } nn{this};
};


int main()
{
  N n;
  N* pn = n.nn.outer_;
    
  assert(&n == pn);
}

[arczi w]

I added an int, a, to the outer structure and wrote
cout << n.nn.outer_->a; in main. The code seems to do it's job but I don't understand what you've written. I have never used the explicit keyword before, not to mention writing such complex-looking statements. Could you please explain what is going on in your work?

[arczi w]

It seems to be the case that for the purpose of displaying an outer variable lines 7, 8 and 9 are non-essential. Moreover, I tried declaring a variable in the inner structure before the pointer. That refuses to work, while declaring a variable after the pointer seems to be legal.

Also: is there a way to make nn a pointer?

[jonnin]

yes, if you put an * on a NN variable you have a pointer.

usually 2 objects do not need to both know about each other in an inheritance chain but sometimes they do. when they do a pointer is an easy way to do it.

struct b; //forward declare lets struct A have a pointer to 'not yet seen' thing b.
struct a
{
int x;
b* bptr;
};

struct b
{
a avar;
}

b bvar;
… code to init everything, a's b pointer is set to parent somewhere in here.
bvar.avar.bptr-> ….

to make all that cleaner you need to construct the a inside of b using b's this pointer to populate a's b pointer. Then you probably only make instances of b class, and it has an a inside that is parental aware.

hopefully that made sense. Im half asleep atm.

[mbozzi]

It seems to be the case that for the purpose of displaying an outer variable lines 7, 8 and 9 are non-essential.

Yes, lines 7-9 are nonessential. Removing them is a compromise for brevity.

I have never used the explicit keyword before

explicit isn't important for the question. Once you look up what it does, just be aware that most single-argument constructors should be explicit. (If you're curious/have questions about it, open a new thread.)

I tried declaring a variable in the inner structure before the pointer. That refuses to work, while declaring a variable after the pointer seems to be legal.

Share the code that doesn't work?

Could you please explain what is going on in your work?

Each object of type N has a member variable of type N::NN named nn. When each object of type N is initialized, the nn's member variable outer_ is set to point to the newly-initialized N.

Also: is there a way to make nn a pointer?

This sounds like an XY problem ( http://xyproblem.info/ ). What issue do you hope to solve by doing this?

[arczi w]

jonnin's answer is what I was looking for. Thank you!