Question

所以这可能是一个简单的修复，但我似乎无法弄明白。

用户输入N输入一个号码。然后用户输入F以将斐波那契序列打印到该数字。

我的原始印刷品0-20（工作代码）被注释掉了。我尝试以相同的方式设置斐波纳契，但在打印数字时会得到垃圾。我现在已将它设置为在for循环之前暂停的位置，以便我可以看到它正在获得的数字。

#include <stdio.h>

char menu();
int read_int(int number);
void display(int answer);

int main(int argc, char ** argv) {
    int number = 0;     
    char choice = 'O';
    while (choice != 'X'){
        choice = menu();    

        if (choice != 'N' && choice != 'F' && choice != 'X') {
            printf("Invalid Input. Enter N, F, or X\n");
        }

        else if (choice == 'N') {
            number = read_int(number);
            printf("\nNumber equals: ");
            printf("%d", number);           
        }

        else if (choice == 'F') {
            //printf("\nNumber equals: ");
            //printf("%d", number);         
            display(number);            
        }
    }
}


char menu () 
{
    char i; 
    printf("\nEnter N to enter an integer from 0 to 20\nEnter F to display the first N+1 numbers (beginning with zero) on the console \nEnter X to quit the program \n");

    printf("Your Choice: ");
    scanf("%s",  &i);
    return i;
}   

int read_int(int number)
{
    number = 0;
    printf("\nEnter an integer 0-20: ");
    scanf("%d", &number);
    if (number >=0 && number <=20)
        return number;
    else{
        printf("Enter a valid number between 0 and 20");    
        return read_int(number);
    }
}

void display(int answer)
{
//  int count = 0;
//  //printf("\nNumber equals: ");
//  //printf("%d", answer); 
//  //int toprint = answer;
//  while (count <= answer){            
//      printf("%d",count);
//      printf(" , ");
//      count=count+1;
   //int n = answer;
   int n = 0;
   n = answer;
   int first = 0, second = 1, next, c;

   //printf("Enter the number of terms\n");


   printf("\nNumber equals: ");
   printf("%d", &answer);
   scanf("%d",&n);
   for ( c = 0 ; c < answer ; c++ )
   {
      if ( c <= 1 )
         next = c;
      else
      {
         next = first + second;
         first = second;
         second = next;
      }
      printf("%d\n",next);
   } 

   //return 0;
}

Answer 1

在您的情况下，问题是从

创建的

 scanf("%s",  &i);

%s需要指向char数组的指针，而不是单个char元素。即使是单个char输入，也会产生内存溢出，导致undefined behaviour。

您可以用

替换它

 scanf(" %c", &i);

另外，printf("%d", &answer);看起来不对，我想你并不是说&会在那里。

通过函数时获取垃圾而不是变量

1 个答案: