c - 验证 C 中的输入

Question

我正在尝试编写一个简单的二进制计算器来重新熟悉 C。出于某种原因，第一次输入验证工作正常，即使数字的第二次验证以几乎相同的方式编写，如果用户输入如果输入错误，while 循环就会无限循环，而不等待新的用户输入。这是代码，感谢您的帮助。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
  char operator[20];
  char valid_operator[4] = "+-*/";
  printf("Enter operator: ");
  scanf("%s", operator);
  printf("You entered: %s\n", operator);
  while(strchr(valid_operator, (int)operator[0]) == NULL) {
    printf("%s is not a valid operator.  Enter +, -, /, or *: ", operator);
    scanf("%s", operator);
  }

代码一直有效到这里。如果用户第一次输入错误的输入，下一部分将陷入无限循环。重新提示永远不会发生。

  int input1;
  int input2;
  printf("Enter the two inputs (separated by whitespace): ");
  int num_ints = 1;
  num_ints = scanf("%d %d", &input1, &input2);
  printf("Input 1: %d.  Input 2: %d.\n", input1, input2);
  while(num_ints < 2){
    printf("Invalid input.  Enter two integers separated by whitespace: ");
    num_ints = 0;
    num_ints = scanf("%d %d", &input1, &input2);
    printf("Input 1: %d.  Input 2: %d.\n", input1, input2);
  }
  return 0;

Answer 1

它无限循环而不等待新用户输入的原因是当scanf时无法读取请求格式的字符(在您的情况下为 %d)它不会前进文件指针，并且在循环的下一次迭代中它将尝试再次读取相同的错误字符.

这与 POSIX 一致:http://pubs.opengroup.org/onlinepubs/009695399/functions/fscanf.html

if the comparison shows that they are not equivalent, the directive shall fail, and the differing and subsequent bytes shall remain unread.

此外，从 man scanf 返回值:

...return the number of input items successfully matched and assigned, which can be fewer than provided for, or even zero in the event of an early matching failure.

所以，你最好结合fgets和sscanf .

do {
    char buf[BUFSZ];
    printf("Enter the two inputs (separated by whitespace): ");
    if(fgets(buf, BUFSZ, stdin) == NULL)
    {
        /* Error exit. */
        break;
    }
    num_ints = sscanf(buf, "%d %d", &input1, &input2);
} while(num_ints != 2);