converting int to pointer - c++

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

Answer 1

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".

Answer 2

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.

Answer 3

#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;
}

Answer 4

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.

Answer 5

(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.

Answer 6

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;
}