在整数用户输入中输入char不会返回错误，而是将其转换为整数？

Question

我写了一个程序返回斐波纳契数列的第n项，n是用户输入的。该程序工作正常，但我输入了一个字母而不是一个整数，看看会发生什么，期待崩溃或错误信息，但它将字母a转换为数字6422368（它将所有我试过的字母转换为相同的数字）。有人能解释为什么发生这种情况吗？

/* Fibonacci sequence that returns the nth term */ 

#include <stdio.h> 

int main() 
{ 
    int previous = 0; // previous term in sequence 
    int current = 1; // current term in sequence 
    int next; // next term in sequence 
    int n; // user input 
    int result; // nth term 

    printf("Please enter the number of the term in the fibonacci sequence you want to find\n"); 
    scanf("%d", &n); 

    if (n == 1) 
    { 
     result = 0; 
     printf("Term %d in the fibonacci sequence is: %d", n, result); 
    } 

    else 
    { 
     for (int i = 0; i < n - 1; i++) // calculates nth term 
     { 
      next = current + previous; 
      previous = current; 
      current = next; 
      if (i == n - 2) 
      { 
       result = current; 
       printf("Term %d in the fibonacci sequence is: %d", n, result); 
      } 
     } 
    } 
} 

Screenshot of Output

Answer 1

当你进入一个非小数字符使用%d，scanf()返回0（错误），并没有设置n。当你打印它时，它是一个非初始化变量，然后打印随机值。

如果要对付这个问题，你可以得到用户的输入为string，检查它是否是一个正确的号码，然后将其转换为int：

#include <math.h> 

int main() 
{ 
    int previous = 0; // previous term in sequence 
    int current = 1; // current term in sequence 
    int next; // next term in sequence 
    int n; // to catch str conversion 
    int i = -1; // for incrementation 
    char str[10]; // user input 
    int result; // nth term 

    printf("Please enter the number of the term in the fibonacci sequence you want to find\n"); 
    scanf("%s", &str); // Catch input as string 
    while (str[++i]) 
     if (!isdigit(str[i])) // Check if each characters is a digit 
      return -1; 
    n = atoi(str); // Convert string to int 

    // Rest of the function