passing pointer to a function

[dwjang]

Hi,

I am completely confused now when I tested the following code.
The last print statement always gave me 0.
Can anyone explain me why?

#include <iostream>

class Stack {
public:
Stack() {next = 0;}
~Stack() {}
int data;
Stack* next;
};

void newStack(Stack* s) {
s = new Stack;
std::cout << "addr of s in newStack function : " << s << std::endl;
}

int main() {

Stack* s = 0;

newStack(s);

std::cout << "addr of s : " << s << std::endl;


return 0;
}

[zb0S216C]

A few things:

1) Don't overload the destructor if it doesn't do anything productive. It's common practice to use the destructor for the deallocation of reserved memory.

2) I see a call to new but no call to delete.

3) Use code tags.

As for your problem, which isn't a problem, s evaluates to the address it points to, which is null (initialisation stage of s), which is effectively the same as 0; thus, std::cout yields 0.

_Wazzak

[vlad from moscow]

Parameter Stack *s declared in your function newStack is a local object that will deleted after execution of the function.

You should redefine your function the following way

Stack * newStack() 
{
   s = new Stack;
   std::cout << "addr of s in newStack function : " << s << std::endl;
   return ( s );
}

and in main use it as

Stack *s = newStack();

[georgewashere]

Stack* s = 0;

Because here you are creating a stack pointer s which points to the value of 0, not to any specific address. When you pass your stack pointer from main to newStack, newStack then assigns the passed pointer to a random place[an address] in memory large enough to hold a Stack object. [this is what happens when you use the new operator]

[dwjang]

Thanks for the quick reply.

What I am confused here is that I am pssing a pointer to a function where a new memory is assigned. So after calling this function, the variable s should have a pointer to that address, right?

Thanks,
dwjang

[zb0S216C]

Yes. new yields a pointer to the allocated memory. s is then assigned to the address where the allocated memory resides. This is called transferring memory ownership, which means giving allocated memory to another pointer.

_Wazzak

[dwjang]

So then, why after calling newStack(Stack* s) function, s is still 0?

[zb0S216C]

I'm not entirely sure, but you can fix that by making s (the parameter) a reference to a pointer (*&) or a pointer-to-a-pointer.

_Wazzak

[dwjang]

I know how to fix this and how to implement more elgant way, but what I don't understand from this simple excise is the concept of passing by reference that I have understood so far.
Passing a pointer to a function should basically be able to change the value of that pointer pointing to. So if I assign a new memory to that passed pointer, I should be able to have a correct pointing address.
Am I missing any fundamental basic concept here?

Thanks,

[vlad from moscow]

You are correctly saying that "Passing a pointer to a function should basically be able to change the value of that pointer pointing to"

But the pointer itself declared as a parameter is a local object of the function. And it is deleted then the function finishes its work.

[dwjang]

As you said, "the pointer itself declared as a parameter is a local object of the function". If this is true, then how can you achieve "pass by reference to the fucntion"?

[zb0S216C]

dwjang wrote:

If this is true, then how can you achieve "pass by reference to the fucntion"?

void newStack( Stack *&In_Pointer ); ?

_Wazzak