Pointer Swap local vs. global func, whats the difference?

[amonsch]

Hi,

does anyone know what the difference is between:

#include <iostream>
 
using std::cout; 
using std::endl;
 
void func( int *p1, int* p2 )
{
    p2 = p1;
}
 
int main(int argc, char* argv[]) {
    int* p2;
    int* p1 = new int;
 
    func(p1, p2);
 
    cout << "P1 Address: " << p1 << " P1 Value: " << *p1 << endl;
    cout << "P2 Address: " << p2 << " P2 Value: " << *p2 << endl;
 
    return 0;
}
 

and

#include <iostream>
 
using std::cout; 
using std::endl;
 
void func( int *p1, int* p2 )
{
    p2 = p1;
}
 
int main(int argc, char* argv[]) {
    int* p2;
    int* p1 = new int;
 
    p2 = p1;
 
    cout << "P1 Address: " << p1 << " P1 Value: " << *p1 << endl;
    cout << "P2 Address: " << p2 << " P2 Value: " << *p2 << endl;
 
    return 0;
}
 

I just wondered because i got different output.

output first:
P1 Address: 0x804b008 P1 Value: 0
P2 Address: 0x804b008 P2 Value: 0

output second:
P1 Address: 0x804b008 P1 Value: 0
P2 Address: 0xbfb680f8 P2 Value: -1078558376

Obviously its not the same address and therefore not the same value, but can someone explain why?

Thanks,
amonsch

P.S.
I use g++ (SUSE Linux) 4.3.2 [gcc-4_3-branch revision 141291]
with openSuse 11.1 on a 32-bit maschine

[mac11]

Firstly I think your outputs are swapped, meaning I think the first program (the one that uses func()) produced the second output.

When you call func() it gets copies of your pointers p1 and p2 and cannot change the actual p1 and p2. Simply put, the assignment inside func() does nothing to the p1 and p2 from main.

When you use assignment (p2 = p1) you're making p2 point to the same place as p1.

I've extended and modified your concept program to demonstrate further

#include <iostream>
 
using namespace std;
 
void func( int *p1, int* p2 )
{
    p2 = p1;
}
 
void ptr_func(int **p1, int **p2)
{
    *p2 = *p1;
}
 
int main(int argc, char* argv[]) 
{
    int* p2;
    int* p1 = new int(3);
 
    cout << "--- before doing anything ---" << endl;
    cout << "p1 @" <<  &p1 << " -> " << p1 << " : " << *p1 << endl;
    cout << "p2 @" <<  &p2 << " -> " << p2 << " : " << *p2 << endl;
    cout << endl;
 
    func(p1, p2);
 
    cout << "--- after using func() ---" << endl;
    cout << "p1 @" <<  &p1 << " -> " << p1 << " : " << *p1 << endl;
    cout << "p2 @" <<  &p2 << " -> " << p2 << " : " << *p2 << endl;
    cout << endl;
 
    p2 = p1;
 
    cout << "--- after using direct assignment ---" << endl;
    cout << "p1 @" <<  &p1 << " -> " << p1 << " : " << *p1 << endl;
    cout << "p2 @" <<  &p2 << " -> " << p2 << " : " << *p2 << endl;
    cout << endl;
 
 
    ptr_func(&p1, &p2);
 
    cout << "--- after using ptr_func() ---" << endl;
    cout << "p1 @" <<  &p1 << " -> " << p1 << " : " << *p1 << endl;
    cout << "p2 @" <<  &p2 << " -> " << p2 << " : " << *p2 << endl;
    cout << endl;
    return 0;
}
 

My output is as follows:
--- before doing anything ---
p1 @0xbff473b0 -> 0x9571008 : 3
p2 @0xbff473b4 -> 0x599ca0 : 0

--- after using func() ---
p1 @0xbff473b0 -> 0x9571008 : 3
p2 @0xbff473b4 -> 0x599ca0 : 0

--- after using direct assignment ---
p1 @0xbff473b0 -> 0x9571008 : 3
p2 @0xbff473b4 -> 0x9571008 : 3

--- after using ptr_func() ---
p1 @0xbff473b0 -> 0x9571008 : 3
p2 @0xbff473b4 -> 0x9571008 : 3


Hopefully this helps you understand pointers and function parameters. Notice that ptr_func() takes the address of p1 and p2, this means that the assignment within ptr_func() actually does modify the p1 and p2 from main.