Acessing a struct after passing to a function

Hello to everyone again.
how to I access my struct variables after i have passed it to a funtion?

I made this struct.

Code:

struct structy
{
	long * one;
	long   two;
	long   Three;
};

using namespace std;

void func1 (structy* structy);

int main()
{
}

void func1 (structy* structy)
	{

	cout<<" the var is " <<structy.one<<endl;//trying to access one here
	
	
	}

This is the error I get.

Code:

	
structhelp.cpp: In function `void func1(structy*)':
structhelp.cpp:29: error: request for member `one' in `structy', which is of non-class type `structy*'

What am I doing wrong?

structy is a pointer, therefor you need to use the -> operater when accessing members, instead of the . operator.

structy->one instead of structy.one

The -> operator is a shortcut for dereferencing (*) and member access (.) combined. Thus, structure->member is the same as (*structure).member.

You could avoid using pointers (in C++ at least) by using a reference.

how to I change the value of what the pointer is pointing to?
In my program I want to use the one variable as a sort of counter. when i try to increment it I am just messing up the the pointer.(I think)

Code:

struct structy
{
	long * one;
	long   two;
	long   Three;
};

using namespace std;

void func1 (structy* structy);

int main()
{
}

void func1 (structy* structy)
	{
                  structy->one++//problem here.  how to I ++ the value
                                          //not the address?
	}

In that code given, you're increment one which is a pointer. Assuming you want to increment the value this is pointing to you have to dereference the pointer as follows:

(*structy->one)++;

Using the magic wand of dereferencing as binky would put it.

The reason for this is that ++ has a higher precedence than *. = and += are lower, so if you don't like the parenthesis, you can use

Code:

*structy->one += 1;

or

Code:

*structy->one = *structy->one + 1;