C++

Forum

 
Functions with Pointers as Parameters

verdastelo (13)
Hi!

My issue with this piece of good is in the function void doit().

The function needs two memory addresses: &foo (address of pointer foo) and goo (which is a pointer).

Why does it work when two integers, instead of addresses, are passed to it?

Here is the raw code:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
 
void DoIt(int &foo, int goo);

int main() {
  int *foo, *goo;
  foo = new int;
  *foo = 1;
  goo = new int;
  *goo = 3;
  *foo = *goo + 3;
  foo = goo;
  *goo = 5;
  *foo = *goo + *foo;
  DoIt(*foo, *goo);
  cout << (*foo) << endl;
}

void DoIt(int &foo, int goo) {
  foo = goo + 3;
  goo = foo + 4;
  foo = goo + 3;
  goo = foo;
}


And here is the code with my comments and cout statements to figure out what is going on:
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
26
27
28
29
30
31
32
33
34
 

void doit (int &foo, int goo) {
  cout << "foo and goo are " << foo << ' ' << goo << endl;
  foo = goo + 3; // Add 5 to *foo. The result is 13.
  cout << "foo and goo are " << foo << ' ' << goo << endl;
  goo = foo + 4; // goo is 13 plus 4. The result is 17.
  cout << "foo and goo are " << foo << ' ' << goo << endl;
  foo = goo + 3; // foo is 12 plus 3. The result is 15.
  cout << "foo and goo are " << foo << ' ' << goo << endl;
  goo = foo; // goo is 15.
}

int main() {
  int *foo, *goo;
  foo = new int; // Allocate some space for an integer pointed to by foo.
  *foo = 1; // The value pointed to by foo is 1.
  cout << "*foo is " << *foo << endl;
  goo = new int; // Allocate some space for an integer pointed to by goo.
  *goo = 3; // The value pointed to by goo is 3.
  cout << "*goo is " << *goo << endl;
  *foo = *goo + 3; // The value pointed to by foo is 6.
  cout << "*foo is " << *foo << endl;
  foo = goo; // goo's address is now stored in foo. foo now points to *goo.
  cout << "foo and goo are " << foo << ' ' << goo << endl;
  *goo = 5; // goo points to 5. So does foo.
  cout << "*goo is " << *goo << endl;
  *foo = *goo + *foo; // foo's new value is 10.
  cout << "*foo is " << *foo << endl;
  cout << "*goo is " << *goo << endl;
  cout << "Now begins the function." << endl;
  doit (*foo, *goo); // *foo and *goo are 10.
  cout << (*foo) << endl; // 20, which is correct.
  return 0;
}
Last edited on
Moschops (7244)
The function needs two memory addresses: &foo (address of pointer foo) and goo (which is a pointer).


No it doesn't. Let's look at the function:
void doit (int &foo, int goo) {

The first parameter, foo, is an int reference.
The second parameter, goo, is an int.

Neither parameter is a pointer. This function doesn't take addresses (pointers) as parameters. That's why it works when you pass it int values. Because that's what these parameters are.
verdastelo (13)
Thank you! The first parameter is not a pointer. What's happening is that foo's value is being passed onto the function by reference.
Moschops (7244)
The second parameter also is not a pointer.
Topic archived. No new replies allowed.