C++

Forum

 
using pointers and structs

kingkong200 (82)
Having trouble understanding using a pointer and structs together and passing through a function. Say for example... 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
 
  using namespace std; 
struct the
{
int a; 
} numbers; 
somefunction (the* pnumbers) 
{
24 = pnumbers.a; // not really sure if that's right.. 
} 

int main ()
{
the *pnumbers = NULL; 
pnumbers = &numbers; 
somefunction (pnumbers); 
return 0; 
}


all in all I am not really sure if I am passing it right in terms of a struct do pass 'the' as the variable definition? More over am I initializing the data correctly in the function?
closed account (28poGNh0)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
 
# include <iostream>
using namespace std;

struct the
{
    int a;
}numbers;

void somefunction(the* pnumbers)// Dont forget the function's type
{
    // 24 = pnumbers.a; // what are you trying to do here?
    pnumbers->a = 24; // because pnumbers is a pointer of a structor,
                      // to access 'a' you should use the arrow operator.

    cout << pnumbers->a << endl;
}

int main ()
{
    the *pnumbers = NULL;
    pnumbers = &numbers;
    somefunction (pnumbers);

    return 0;
}

Last edited on
kingkong200 (82)
so will 
1
2
 
 
pnumbers ->a =24;


be equal to 
1
2
3
4
 
int main () 
{ 
numbers.a = 24; 
}


because in essence that is what I am trying to do but within a function and keeping it memory efficient.
closed account (28poGNh0)
No It will be equal to

(*pnumbers).a = 24
Last edited on
Codeez (86)
@kingkong200

-> is an alternate way for do what techno01 just wrote. You're dereferencing the pointer to get the struct, and as normal you use the . to get the member of that struct. We use -> simply because the other way is ugly. The ( ) are for precedence and * for dereference.
Last edited on
closed account (28poGNh0)
quite yes also you should know that
(*ptr).member : is c style
and ptr->member : is c++ style
kingkong200 (82)
sorry, what I getting confused about is when you initialized 
1
2
 
 
pnumbers = &numbers;


I thought essentially you are controlling the numbers struct and not entirely different struct called pnumbers? or perhaps I was wording it wrong above what I meant is 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
 
 
# include <iostream>
using namespace std;

struct the
{
    int a;
}numbers;

void somefunction(the* pnumbers)
{
    pnumbers->a = 24;

    cout << pnumbers->a << endl;
}

int main ()
{
    the *pnumbers = NULL;
    pnumbers = &numbers;
    somefunction (pnumbers);
    cout << numbers.a << endl;  // Will this be 24? such are my intentions... 

    return 0;
}
Last edited on
closed account (28poGNh0)
sorry, what I getting confused about is when you initialized
pnumbers = &numbers;


first I did not initialized... you did I just copied your code :)

second dont be confused because we will start from 0

well you declared an instance(numbers) for the structor "the"
then you declared a structor pointer(pnumbers) that points to this object(numbers)

pnumbers = &numbers means that we will take the address of the object numbers ,assign it to pnumbers structor pointer ,so that pnumbers can points all the time at the object numbers

In the function "someFunction" we have an
entirely different struct called pnumbers
but since it take the address of the pnumbers It can access all numbers members ,as I did pnumbers->a = 24;

finaly in main function you can write cout << numbers.a << endl; and it will be 24

also this statement pnumbers ->a =24;

equals to 

1
2
3
4
 
int main () 
{ 
numbers.a = 24; 
}

My mistake I did not pay attention I thought you did put numbers.a in "someFunction" function sorry :(
Last edited on
kingkong200 (82)
Awesome Thanks.
Topic archived. No new replies allowed.