pass by pointer and pass a NULL

[csxpeng]

Hi all, in below test code, if I pass a NULL pointer to the add(int*) function, I still couldn't get the allocated space in my main function. And it thus gave me segmentation fault.

Could you advise why? A workaround for this issue would be pass by a reference of a point, eg,

void add(int* &i)

Is there any other solutions? Thanks very much.

#include <iostream>

using namespace std;

void add(int* i) {
  if(i==NULL) {
    i=new int(0);
    (*i)++;
    return;
  }

  (*i)++;
}

int main(int argc, char **argv)
{
  int* i=NULL;
  add(i);

  cout << *i << endl;

  return 0;
}

[Duthomhas]

Your function header is different than what you list.

void add(int* &i)

This is a reference to a pointer-to-int.

void add(int* i)

This is just a pointer-to-int (which is passed by value).

Hence, your function cannot modify the i in main() (declared on line 17). If you fix your add() function to take the proper type of argument, it will work.

You should also remember to delete i; before you return 0;.


One last thing: The name "add" is not properly informative. Use "increment" or "inc" or "incr" or "bump" etc.

Hope this helps.

[csxpeng]

Hi Duoas, your reply is very helpful. Thanks. So what you mean is the function cannot modify i (the pointer itself) in main(). However if I pass a non-NULL pointer, and I can get the updated value pointed by i.

Maybe I mixed two concepts here, the pointer itself and the value pointed by the pointer? Thanks.

#include <iostream>

using namespace std;

void inc(int* i) {
  if(i==NULL) {
    i=new int(0);
    (*i)++;
    return;
  }

  (*i)++;
}

int main(int argc, char **argv)
{
  int* i=new int(10);

  inc(i);
  cout << *i << endl;

  delete i;

  return 0;
}

[IceThatJaw]

If this helps,

the pointer itself is just an address.

cout << i << endl; will just cout the address or 0 if it is set to NULL.

When you write *i you are referring to the value pointed to by i.


And just to be safe, I always check if a pointer is NULL before I dereference it by writing if (i)

This will return false if the pointer i is set to NULL. Dereferencing a NULL pointer will result in a run-time seg fault. They are not fun.

[firedraco]

Yes.

On line 7, you are modifying the pointer i.
On line 8/12, you are modifying the value that i is pointing at.

They are different.

[Duthomhas]

...which is what he intended, of course. The problem is that he is confused by combining the ideas between pointers and references and the redirections complicit in each...

#include <iostream>

using namespace std;

void inc(int* &i) {
  if(i==NULL) {
    i=new int(0);
    (*i)++;
    return;
  }

  (*i)++;
}

int main(int argc, char **argv)
{
  int* i=NULL;

  inc(i);
  cout << *i << endl;

  inc(i);
  cout << *i << endl;

  return 0;
}

Hope this helps.