Pointer Problem

Esam · #1 22-May-2006, 14:51

I'm sure this is really simple and stupid, but I'm not exactly sure what I am doing wrong. I have an array in an object. I would like to accept a pointer as an argument and set that pointer to the address of some index in that array for the calling class to use. Basically, the pointer is always equal to 0 when I try to use it. I probably posted more than necessary but it is very simple so here is a test example that exhibits the problem. The main and setptr functions should make it obvious what I am trying to do.

CPP / C++ / C Code:

#include <iostream>
using namespace std;

class ptr_test
{
public:
   ptr_test();
   ~ptr_test();
   
   void setptr(int * ptr);

private:
   int * testptr;
};

ptr_test::ptr_test()
{
   testptr = new int[20];
}

ptr_test::~ptr_test()
{
   if (testptr)
   {   
      delete [] testptr;
      testptr = 0;
   }
}

void ptr_test::setptr(int * ptr)
{ 
   ptr = testptr;
   cout << "Member: " << testptr << endl;
   cout << "Argument: " << ptr << endl;
}

int main(int argc, char *argv[])
{
   ptr_test p;
   int * int_ptr;
   int_ptr = 0;
   cout << "Initial: " << int_ptr << endl;
   p.setptr(int_ptr);
   cout << "Set: " << int_ptr << endl;    
}

The result is as follows:
Initial: 0
Member: 0x3d3ca8
Argument: 0x3d3ca8
Set: 0

It is like it goes out of scope or something but I don't think it should. Any answers?

Thanks

davekw7x · #2 22-May-2006, 15:39

Quote:

Originally Posted by Esam

but I'm not exactly sure what I am doing wrong.

If you want a function that changes the value of one of its arguments (so that the change is reflected back into the calling program), there are a couple of ways:

1. Make the argument a pointer to the variable, then in the function de-reference the pointer to access the value of the variable (this works for C and C++). (So, if the variable is a pointer (int *arg), you would pass a pointer to the pointer (int **arg).) Syntax in the calling program changes to use the address of the pointer. Syntax in the function changes to dereference the pointer-to-pointer.

2. Make the argument a reference to the variable, then whatever you do in the function is actually acting on the original (this works for C++ only).

(Otherwise the function only changes the value of its local copy of the argument, and this does not change the value of the corresponding variable in the calling program.)

For C++, many people prefer the second one:

CPP / C++ / C Code:

   void setptr(int * &ptr);// ptr is a reference to a variable (the variable is a pointer to int)
.
.
.
void ptr_test::setptr(int * &ptr)
{ 
   ptr = testptr;
   cout << "Member: " << testptr << endl;
   cout << "Argument: " << ptr << endl;
}
.
.
.
   p.setptr(int_ptr);
.
.
.

With these changes (and only these changes), I saw the following output:

Code:

Initial: 0
Member: 0x6a2268
Argument: 0x6a2268
Set: 0x6a2268

Regards,

Dave

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Pointer Problem

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