pointers question

[adam2016]

Hi I am in the process of figuring out linked lists but one thing that is catching me out and I know it's probably fairly basic by how come when I try to create a new ship object and set it equal to getNewShip() I get an error?

#include <iostream>

using namespace std;


struct ship{

    int x_coOrd;
    int y_coOrd;
    ship* nextShip;

};


ship* getNewShip(){


     ship* p_ship = new ship;
     p_ship->x_coOrd = 0;
     p_ship->y_coOrd = 0;
     p_ship->nextShip = p_ship;

     return p_ship;

}



int main()
{
   ship one = getNewShip(); // THIS DOES NOT WORK


}

but when i set ship one to be a pointer instead it compiles and works

#include <iostream>

using namespace std;


struct ship{

    int x_coOrd;
    int y_coOrd;
    ship* nextShip;

};


ship* getNewShip(){


     ship* p_ship = new ship;
     p_ship->x_coOrd = 0;
     p_ship->y_coOrd = 0;
     p_ship->nextShip = p_ship;

     return p_ship;

}



int main()
{
   ship* one = getNewShip(); // THIS WORKS


}

[cire]

I'm not sure whats so confusing. getNewShip returns a pointer and in the first snippet one is not a pointer. What should a fully constructed object do with a pointer to another object? What does a house do with the address to another house?

[keskiverto]

An another "interesting" thing occurs on line 22.
We can see what happens there, but would you explain it (and why you do that)?

[adam2016]

yeah it's pretty straight forward but what I don't understand is getNewShip returns a pointer

then ship* one a pointer itself stores a pointer,I thought a pointer can only store a memory address,so ship one would store a pointer that points to memory in the heap wouldn't you need a pointer that stores a pointer like **p?

@Kesk at line 22 a pointer stores the memory address of that current ship it's supposed to be set to NULL because I have not made a head pointer yet so yeah that should be null.

[cire]

then ship* one a pointer itself stores a pointer,I thought a pointer can only store a memory address,so ship one would store a pointer that points to memory in the heap wouldn't you need a pointer that stores a pointer like **p?

A pointer to a pointer also stores only an address.

#include <iostream>

int main()
{
    int value = 73;             // an int object.

    int* ptr_a = &value;        // pointer-to-an-int.  Stores the address an int is stored at.

    int* ptr_b = ptr_a;         // pointer-to-an int.  Stores the address an int is stored at.

    int** ptr_c = &ptr_a;       // pointer-to-a-pointer-to-int.  Stores the address a pointer is stored at.


    std::cout << "value: " << value << "\t\t&value: " << &value << '\n';
    std::cout << "ptr_a: " << ptr_a << "\t\t&ptr_a: " << &ptr_a << '\n';
    std::cout << "ptr_b: " << ptr_b << "\t\t&ptr_b: " << &ptr_b << '\n';
    std::cout << "ptr_c: " << ptr_c << "\t\t&ptr_c: " << &ptr_c << '\n';
}

value: 73               &value: 000000CC64CFFDE0
ptr_a: 000000CC64CFFDE0         &ptr_a: 000000CC64CFFDE8
ptr_b: 000000CC64CFFDE0         &ptr_b: 000000CC64CFFDF0
ptr_c: 000000CC64CFFDE8         &ptr_c: 000000CC64CFFDF8

You can see that ptr_a and ptr_b store the same address, while residing at different ones. ptr_c stores the address of ptr_a while residing at yet another address in memory.

A pointer-to-pointer is only useful when you have a pointer hanging around in memory somewhere and you want to store its address (a situation that is not present in the OP.)

[adam2016]

thanks cire that makes a little more sense =)

just a follow up when I try to do this how come the compiler throws an error?

int a = 7;
int *b = &a;
int *c = &b;

but when I do this,it works fine

int a = 7;
int *b = &a;
int *c = b;

[keskiverto]

Address of object, where the type of object is T, must be stored in a "pointer to T" (aka T*).

The a is of type int and thus you store its address in int *
The b is of type int* and thus you must store its address in int* *

The b holds the address of a. Therefore:

c = b;  // copy address of a to c
c = &b; // copy address of b to c 

[adam2016]

Thanks guy I understand it now,seems so complicated yet in reality it's pretty straight forward =)

Thanks