converting int to pointer

Question

I want to save int value to a pointer variable. But I get an error:

#include <iostream>
using namespace std;

int main()
{
  int *NumRecPrinted = NULL;
  int no_of_records = 10;
  NumRecPrinted = (int*)no_of_records; // <<< Doesn't give value of NumRecPrinted

  cout << "NumRecPrinted!" << NumRecPrinted;
  return 0;
}

I tried doing this but I get 0 as return:

int main()
{
    int demo(int *NumRecPrinted);
    int num = 2;
    demo(&num);
    cout << "NumRecPrinted=" << num;    <<<< Prints 0
    return 0;
}

int demo (int *NumRecPrinted)

{
    int no_of_records = 11;
    NumRecPrinted = &no_of_records;
}

NumRecPrinted returns as 0

Why would you want to do that? There are few good reasons to convert an integer into a pointer, so I am wondering whether what you are trying to do might not be better achieved somehow else. Or do you mean *obtain a pointer to an integer?* — David Rodríguez - dribeas, Feb 09 '12 at 14:21
i am doing this inside a function which needs to return the pointer. — AJ., Feb 09 '12 at 14:24
In your `demo` function, you modify `NumRecPrinted` to point to a local variable... It will not be useful outside `demo`... — Abrixas2, Feb 09 '12 at 14:31
@AJ. I hope you're not planning [a hotel heist](http://stackoverflow.com/a/6445794/46642). — R. Martinho Fernandes, Feb 09 '12 at 14:36

score 10 · Answer 1 · answered Feb 09 '12 at 14:09

It's sometimes useful to "encode" a non-pointer value into a pointer, for instance when you need to pass data into a pthreads thread argument (void*).

In C++ you can do this by hackery; C-style casts are an example of this hackery, and in fact your program works as desired:

#include <iostream>
using namespace std;

int main()
{
  int *NumRecPrinted = NULL;
  int no_of_records = 10;
  NumRecPrinted = (int*)no_of_records;

  cout << "NumRecPrinted!" << NumRecPrinted; // Output: 0xa (same as 10)
  return 0;
}

You just need to realise that 0xa is a hexadecimal representation of the decimal 10.

However, this is a hack; you're not supposed to be able to convert ints to pointers because in general it makes no sense. In fact, even in the pthreads case it's far more logical to pass a pointer to some structure that encapsulates the data you want to pass over.

So, basically... "don't".

But if you must ... use [intptr_t](http://stackoverflow.com/questions/6326338/why-when-to-use-intptr-t-for-type-casting-in-c) — Useless, Feb 09 '12 at 14:24
... or at least `static_assert(sizeof(int) >= sizeof(void*), "the size of int must be greater than or equal to the size of a pointer");`, i.e., document your assumptions. — R. Martinho Fernandes, Feb 09 '12 at 14:34

mattjgalloway · Accepted Answer · 2012-02-09T14:11:14.400

5

You want to be doing this:

NumRecPrinted = &no_of_records;

i.e. you're taking the address of no_of_records and assigning it to NumRecPrinted.

And then to print it:

cout << "NumRecPrinted!" << *NumRecPrinted;

i.e. you're dereferencing NumRecPrinted which will get the int stored at the memory address pointed to by NumRecPrinted.

edited Feb 09 '12 at 14:11

answered Feb 09 '12 at 14:02

mattjgalloway

34,562
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score 3 · Answer 3 · answered Feb 09 '12 at 14:42

#include <iostream>
using namespace std;

int main()
{
int *NumRecPrinted = NULL; // assign pointer NumRecPrinted to be valued as NULL
int *NumRecPrinted2 = NULL;
int no_of_records = 10; // initialize the value of the identificator no_of_records 
NumRecPrinted = (int*)no_of_records; // sets a pointer to the address no_of_records
NumRecPrinted2 = &no_of_records; // gives a pointer to the value of no_of_records

cout << "NumRecPrinted!" << NumRecPrinted;  // address of no_of_records 0000000A
cout << "NumRecPrinted!" << *NumRecPrinted2; // value of no_of_records 10
system("pause"); // ninja 
return 0;
}

Why no indentation? And why did you add `system("pause")`? Doing that makes baby jesus cry. — Lightness Races in Orbit, Jan 28 '13 at 13:24

score 1 · Answer 4 · answered Feb 09 '12 at 14:02

Here is the corrected version:

#include <iostream>
using namespace std;
int main()
{
  int *NumRecPrinted = NULL;
  int no_of_records = 10;
  NumRecPrinted = &no_of_records; // take the address of no_of_records

  cout << "NumRecPrinted!" << *NumRecPrinted; // dereference the pointer
  return 0;
}

Note the added ampersand and the asterisk.

score 1 · Answer 5 · answered Feb 09 '12 at 14:04

1

(int *)no_of_records gives you a pointer to the address no_of_records. To get a pointer to the value of no_of_records, you need to write &no_of_records.

answered Feb 09 '12 at 14:04

Abrixas2

3,157
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score 0 · Answer 6 · answered Feb 09 '12 at 14:21

0

I really like using union for this sort of stuff:

#include <iostream>
using namespace std;

int main()
{
  static_assert(sizeof(int) == sizeof(int*));

  union { int i; int* p; } u { 10 };

  cout << "NumRecPrinted! " << u.p;
  return 0;
}

answered Feb 09 '12 at 14:21

Gigi

4,876
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That is undefined behavior in C++. – Julius May 13 '19 at 20:52

converting int to pointer

6 Answers6