Question

我在各个论坛上都试图了解这个问题。我无法完全理解这个问题以及为什么我找不到解决方案的原因是因为我对C ++很新，我不理解错误信息。

这是我在C ++中的代码，它通过排列或组合公式查找可能性的数量。每当我尝试编译并运行时，我都会收到消息：

在0x6a8613af（msvcr100d.dll）中的第一次机会异常 Combinations_Permutations.exe：0xC0000005：访问冲突读取位置0x00000005。 0x6a8613af处于未处理的异常（msvcr100d.dll）在Combinations_Permutations.exe中：0xC0000005：访问冲突读取位置0x00000005。

我在很多其他论坛上都知道“访问冲突读取位置0x00 ......”肯定可以指示空指针。但我无法看到我遇到这样一个空问题。也许我的变量是全局访问的，它们不是YET初始化的？这是我的代码，我已经有一段时间了...就像我说我很新。所以请告诉我我的错误。谢谢。

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);
    }
}

Answer 1

这是一个错误：

printf("Number of possibilities: %s", solution);

solution是int，不是以空字符结尾的字符串：使用%d。

使用类型安全的std::cout代替printf()，可以防止出现此错误：

std::cout << "Number of possibilities: " << solution;

Answer 2

有问题的一行是：

printf("Number of possibilities: %s", solution);

您告诉printf solution是char*，因此它会尝试取消引用(char*)solution来打印“C字符串”的内容（大概是如果您的特定错误消息，solution的值为5。

将%s更改为%d，或使用std::cout代替printf以获得类型安全，并首先避免此类问题。

完成初学者：空指针问题？

2 个答案: