完全初学者:空指针问题

我浏览了所有论坛，试图了解这个问题。我不能完全理解这个问题，也不能找到解决方案的原因是，我对C++还很陌生，不理解错误消息。

这是我在C++中的代码，它从排列或组合公式中找到了可能性的数量。每次我尝试编译和运行时，我都会收到这样的消息：

中0x6a8613af（msvcr100d.dll）处的首次机会异常Combinations_Permutations.exe:0xC0000005:读取访问冲突位置0x00000005。0x6a8613af处出现未处理的异常（msvcr100d.dll）在Combinations_Permutations.exe:0xC0000005:读取访问冲突位置0x00000005。

我在许多其他论坛上了解到，"访问违规读取位置0x00…"肯定表示空指针。但我看不出我在哪里遇到这样一个无效的问题。也许我的变量正在被全局访问，而它们并没有被YET初始化？这是我的代码，我已经用了一段时间了。。。就像我说的，我还是个新手。所以请把我的错误告诉我。非常感谢。

我的代码：

    #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);
    }
}