C++ program to convert decimal to binary is getting segmentation fault - c++

Question

This is a c++ program to convert decimal number to binary number. Well there are many possible ways to implement this but as I learned about the static variable I thought to make a use of it. So the program is
#include <iostream>
using namespace std;
int **binary(int num, int &k) {
static int *p;
int i = 0;
while (num > 0) {
*p = num % 2;
p++;
num = num / 2;
k++;
}
return &p;
}
int main() {
int n;
int k = 0;
cout << "\n Enter the number to be converted into binary : ";
cin >> n;
int **ptr;
ptr = binary(n, k);
cout << "\n The number of bytes in the binary number is : " << k << endl;
cout << "\n The binary code is : \n";
for (int i = 0; i < k; i++)
cout << **(ptr+i);
return 0;
}
Output:
Enter the number to be converted into binary : 33
Segmentation fault
After debugging this code I'm getting seg fault at line 9:9
i.e
*p = num % 2;
I don't know why this is leading me to access an unoccupied memory space in the stack.

Answer 1

The reason there is a crash is that p is initialized to nullptr. Dereferencing an uninitialized variable is undefined behavior, hence you get the crash.
Disclaimer: The following code is intended to fix your code. It is not intended to illustrate a proper way of doing this, or show a good way to code in general.
If you wish to play with function-static variables, and your function must return a pointer to pointer, initialize p by pointing it to another static variable which provides you with a buffer:
static int pVal[100];
static int * p;
p = pVal; // This should be done in an assignment, not in initializer
Note: Returning pointers to static storage makes your code non-reentrant, which is generally a very bad practice. This code is fine as a learning exercise, but it's not something one should use in production code.

Answer 2

static int * p; is never set. It is initialized to NULL, dereferencing it invokes undefined behavior. Using a static buffer is not recommended, it is better to pass an array to the function and return the number of bits. Furthermore, you should output these bits in the reverse order.
Since you are interested in static local variables, here is a corrected version of your code:
#include <iostream>
using namespace std;
int *binary(int num, int &k) {
static int bits[sizeof(int) * 8];
int *p = bits;
k = 0;
while (num > 0) {
*p = num % 2;
p++;
num = num / 2;
k++;
}
return p;
}
int main() {
int n;
int k = 0;
cout << "\n Enter the number to be converted into binary : ";
cin >> n;
int *ptr = binary(n, k);
cout << "\n The number of bits in the binary number is : " << k << endl;
cout << "\n The binary code is : \n";
for (int i = 0; i < k; i++)
cout << ptr[k - i];
return 0;
}
Here is a version that does not use a static buffer:
#include <iostream>
using namespace std;
int binary(unsigned int num, int *dest) {
for (int i = 0;;) {
dest[i++] = num & 1;
if ((num >>= 1) == 0)
return i;
}
}
int main() {
unsigned n;
int bits[sizeof(n) * 8];
cout << "\nEnter the number to be converted into binary: ";
cin >> n;
int k = binary(n, bits);
cout << "\nThe number of bits in the binary number is: " << k << endl;
cout << "\nThe binary code is: ";
for (int i = k; i-- > 0;)
cout << bits[i];
cout << endl;
return 0;
}

Answer 3

The code has numerous problems:
binary() does not initialise the pointer p to anything.
binary() dereferences the initialised p and modifies the value.
The conversion writes from LSB to MSB and the outputs writes in that order too.
You need to allocate a buffer to receive the conversion. This is safest done by the caller. Since ultimately you output the characters 1 or 0, your conversion may as well write those character values directly to a string.