Question concerning vector of pointers

[Student555]

Why does the following print out different memory addresses for the bullet object

  	vector<Bullet*> Gun;
	Gun.push_back(&bullet);
	Gun.push_back(&bullet);
	Gun.push_back(&bullet);
	Gun.push_back(&bullet);

	const Bullet * ptr = &bullet;
	int count = 0;

	cout << "Original bullet object: " << ptr << endl;
	cout << "bullet object address: " << &bullet << endl;
	cout << "  " << endl;
	cout << "  " << endl;

	for(vector<Bullet>::const_iterator itr = Gun.begin(); itr != Gun.end(); itr++)
	{
		count++;
		cout << "Address " << count << " :" << &(*itr) << endl;
	}

shouldn't the pointers in the vector be printing out the same address for the bullet object. Why are a different set of memory addresses allocated?

[Bdanielz]

I think, You are printing out an iterator in your loop, not the bullet object address.
Also I cannot get your code to run. What are you trying to accomplish that is not working?

[dhayden]

Your code doesn't compile. Line 15 declares itr as an iterator for vector<Bullet> but Gun is a vector<Bullet*>. This code works:

#include <iostream>
#include <vector>

using std::cout;
using std::vector;
using std::endl;

typedef int Bullet;

int main()
{
    int bullet;
    vector<Bullet*> Gun;
    Gun.push_back(&bullet);
    Gun.push_back(&bullet);
    Gun.push_back(&bullet);
    Gun.push_back(&bullet);

    const Bullet * ptr = &bullet;
    int count = 0;

    cout << "Original bullet object: " << ptr << endl;
    cout << "bullet object address: " << &bullet << endl;
    cout << "  " << endl;
    cout << "  " << endl;

    for(vector<Bullet*>::const_iterator itr = Gun.begin(); itr != Gun.end(); it\
r++)
        {
            count++;
            cout << "Address " << count << " :" << *itr << endl;
        }
}

$ ./foo
Original bullet object: 0x22caac
bullet object address: 0x22caac


Address 1 :0x22caac
Address 2 :0x22caac
Address 3 :0x22caac
Address 4 :0x22caac

[Student555]

@Bdanielz
Doesn't an iterator act like a pointer? Shouldn't the iterator be printing out the exact address as the bullet object?

EDIT:
Did not know you responded dhayden. Your code let me to the right outout but I'm left with a question

what is the difference between

cout << "Address " << count << " :" << &(*itr) << endl;

and

cout << "Address " << count << " :" << *itr << endl;

The first gives different memory addresses for the pointers while the second one prints the same memory address, what was expected.

[Bdanielz]

Doesn't an iterator act like a pointer?

I think a pointers and iterators are closely related but not the same.

Shouldn't the iterator be printing out the exact address of the bullet object?

It appears you were correct and I was wrong. I could not test due to not being able to compile your code. In other words I guessed.

[cire]

what is the difference between

cout << "Address " << count << " :" << &(*itr) << endl;

This prints out the address of whatever is referenced by iter.

and cout << "Address " << count << " :" << *itr << endl;

Prints out the value of whatever is referenced by itr. For a pointer, the value is an address (typically of another object.)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

#include <iostream> int main() { int val = 6; int* val_ptr = &val; // We don't expect the value contained by val_ptr to be 6 std::cout << "val_ptr: " << val_ptr << '\n'; // similarly we expect that the object val_ptr will have an address // that is different than val. std::cout << "&val_ptr: " << &val_ptr << '\n'; // The same goes for pointers. int * ptr; int ** ptr_ptr = &ptr; std::cout << "ptr: " << ptr << '\n' ; // some unspecified value std::cout << "ptr_ptr: " << ptr_ptr << '\n' ; // the address of ptr std::cout << "&ptr_ptr: " << &ptr_ptr << '\n'; // the address of ptr_ptr }

val_ptr: 000000D6C4EFFB84 &val_ptr: 000000D6C4EFFBA8 ptr: CCCCCCCCCCCCCCCC ptr_ptr: 000000D6C4EFFBC8 &ptr_ptr: 000000D6C4EFFBE8

[Student555]

So to recap

pointers, in addition to the pointed to value and address, also have a memory address of their own. Since they are objects that are allocated on the stack.

[AbstractionAnon]

Since they are objects that are allocated on the stack.

Pointers do not have to be on the stack. They can also be on the heap.

[Student555]

Pointers do not have to be on the stack. They can also be on the heap.

Right in the following sense:

Object * ptr = new Object(); // Though I understand this is no longer recommended  

I meant in the regular raw pointer sense:

int * ptr;

[Moschops]

Both this pointer:
Object * ptr = new Object(); // Though I understand this is no longer recommended
and this pointer:
int * ptr;
are on the stack.

[Student555]

Object * ptr

is on the stack, but

new Object()

is on the heap. The Object * ptr is an object allocated on the stack but points to heap memory correct?

[AbstractionAnon]

That's correct.

Consider the following:

  int *ptr = new int * [100];

That just created 100 int pointers on the heap. In this example, ptr is on the stack, but it does not have to be.

[Student555]

That's clear Anon.
Thanks for the help everyone.