Complete Beginner: Null pointer issue? - c++

Question

I have looking all over forums to try and understand this issue. The reasons I cannot fully understand the issue and why I cannot find a solution is because I am fairly new with C++ and I do not understand the error message.
This is my code in C++ that finds the number of possibilities from permutation or combination formulas. Every I try and compile and run, I get messages that say:
First-chance exception at 0x6a8613af (msvcr100d.dll) in
Combinations_Permutations.exe: 0xC0000005: Access violation reading
location 0x00000005. Unhandled exception at 0x6a8613af (msvcr100d.dll)
in Combinations_Permutations.exe: 0xC0000005: Access violation reading
location 0x00000005.
I've learned on many other forums that the "access violation reading location 0x00..." could definitely indicate null pointer. But I cannot see where I encounter such a null issue. Maybe my variables are being accessed globally, where they are not YET initialized?
Here is my code, I have been at it for a while... like I said I'm fairly new. So please inform me of my mistake(s). Thank you.
My code:
#include <iostream>
#include "conio.h";
using namespace std;
int run_combination(int n, int r);
int run_permutation(int n, int r);
int solve_factorial(int f);
int f_value = 1; //factorial value used recursively
int n_input, r_input;
char choice;
char order;
void main(){
//if user types choice as 'q', while loop ends
while(choice != 'q'){
printf("How many values? (1-9) \n");
printf("User: ");
cin >> n_input;//user input for variable n
printf("n_input: %i", n_input);
printf("\nHow many will be chosen at a time out of those values? (1-9)\n");
printf("User: ");
cin >> r_input; //user input for variable r
printf("\nDoes order matter? (y/n)\n");
printf("User: ");
cin >> order; //'y' if order is taken into consideration(permutation)
//'n' if order it NOT taken into consideration(combination)
int solution = 0; //temporary variable that represents the solution after running
//n and r through the permutation or combination formula
//if user input values for n and r are in between 1 and 9, then run
//combination or permutation
if (n_input <= 9 && n_input >= 1 && r_input <= 9 && r_input >= 1){
if (order == 'y')
solution = run_permutation(n_input, r_input);
else if (order == 'n')
solution = run_combination(n_input, r_input);
else
printf("\nError. Please type 'y' or 'n' to determine if order matters.\n");
//if n < r, run_permutation or run_combination returns 0
if (solution == 0){
printf("Error! You can't choose %i values at a time if there \n",
"are only %i total values. Type in new values next loop \n.", r_input, n_input);
}
else
printf("Number of possibilities: %s", solution);
}
else{ //else error message if numbers are out of range...
printf("Next loop, type in values that range from 1 to 9.\n");
}
//option 'q' to quit out of loop
printf("Type 'q' to quit or enter any key to continue.\n");
printf("User: ");
cin >> choice;
}
_getch();
}
/*
Returns solved combination of parameters n and r
Takes the form: n! / r!(n-r)!
*/
int run_combination(int n, int r){
if (n < r) //solution is impossible because you can't choose r amounnt at a time if r is greater than n
return 0;
int n_fac = solve_factorial(n); //n!
int r_fac = solve_factorial(r); //r!
int nMinusr_fac = solve_factorial(n-r); //(n-r)!
int solve = ((n_fac) / ((r_fac)*(nMinusr_fac))); // plugging in solved factorials into the combination formula
return solve;
}
int run_permutation(int n, int r){
if (n < r)
return 0;
int n_fac = solve_factorial(n);
int nMinusr_fac = solve_factorial(n-r);
int solve = ((n_fac) / (nMinusr_fac)); //plugging in factorials into permutation formula
return solve;
}
int solve_factorial(int f){
if (f-1==0 || f == 0){ //if parameter f is 1 or 0, return 1
int temp = f_value;
f_value = 1; //reset f_value so that f_value remains 1 at the start of every new factorial
return temp;
}
else{ //else multiply f_value by f-1
f_value *= f;
return solve_factorial(f-1);
}
}

Answer 1

This is a mistake:
printf("Number of possibilities: %s", solution);
solution is an int, not a null terminated string: use %d.
Using std::cout which is typesafe, instead of printf(), would have prevented this error:
std::cout << "Number of possibilities: " << solution;

Answer 2

The problematic line is:
printf("Number of possibilities: %s", solution);
You're telling printf that solution is a char*, and so it tries to dereference (char*)solution to print the contents of the "C-string" (presumably when solution has the value 5 in the case of your particular error message).
Change %s to %d, or use std::cout instead of printf to gain type-safety and avoid this sort of issue in the first place.