在整數用戶輸入中輸入char不會返回錯誤，而是將其轉換爲整數？

Question

我寫了一個程序返回斐波納契數列的第n項，n是用戶輸入的。該程序工作正常，但我輸入了一個字母而不是一個整數，看看會發生什麼，期待崩潰或錯誤信息，但它將字母a轉換爲數字6422368（它將所有我試過的字母轉換爲相同的數字）。有人能解釋爲什麼發生這種情況嗎？

/* 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

當你進入一個非小數字符使用%d，scanf()返回0（錯誤），並沒有設置n。當你打印它時，它是一個非初始化變量，然後打印隨機值。

如果要對付這個問題，你可以得到用戶的輸入爲string，檢查它是否是一個正確的號碼，然後將其轉換爲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