完整初学者：空指针的问题？

Question

我已经找遍了所有的论坛来尝试和理解这个问题。我无法完全理解问题的原因以及为什么找不到解决方案是因为我对C++相当陌生，而且我不明白错误消息。

这是我在C++中的代码，它可以从排列或组合公式中找到可能的数目。每次我尝试编译和运行，我得到消息说：

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.

我已经在许多其他论坛的“访问冲突读取地址0x00 ......”肯定可以表明空指针教训。但我看不到我遇到这样一个空问题。也许我的变量正在全球访问，他们不是初始化？ 这是我的代码，我一直在这里一段时间...就像我说我是相当新的。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

问题的行是：

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

你告诉printf即solution是char*，因此它试图取消引用(char*)solution要打印的内容“C字符串”（假设您的特定错误消息为solution的值为5）。

变化%s到%d，或使用std::cout代替printf获得类型安全和避免这类问题摆在首位。

Answer 2

这是一个错误：

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

solution是int，不是一个空结束的字符串：使用%d。

使用std::cout这是类型安全，而不是printf()，就可以防止此错误：

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