C/C++ 用feof函数，输出时怎么把最后的一个字符重复输出了？

Question

C版 ：

#include <stdio.h>
#include <errno.h>
#include <ctype.h>
int main(int argc,char **argv) {
 FILE *input = fopen("in.test.txt","r");
 char c = '\0';
 while (! feof(input)) {
  fscanf(input,"%c",&c);
  printf("%c",tolower(c));
 }
 printf("\b\n");
 return 0;
}
/*
 * in.test.txt: "XXX"
 * OutPut: "xxxx"
*/

C++版 ：

#include <cstdio>
#include <cerrno>
#include <cctype>
using namespace std;
int main(int argc,char **argv) {
 FILE *input = fopen("in.test.txt","r");
 char c = '\0';
 while (! feof(input)) {
  fscanf(input,"%c",&c);
  printf("%c",tolower(c));
 }
 printf("\b\n");
 return 0;
}
/*
 * in.test.txt: "XXX"
 * OutPut: "xxxx"
*/

求助大佬们为什么会最后一个字符输出两次?

Answer 1

#include <cstdio>
#include <cerrno>
#include <cctype>
using namespace std;
int main(int argc,char **argv) {
 FILE *input = fopen("c:\\in.test.txt","r");
 char c = '\0';
 while (1) {
  if (fscanf(input,"%c",&c) == EOF) break;
  printf("%c",tolower(c));
 }
 printf("\b\n");
 return 0;
}

scanf失败，c还是之前的值，所以多输出1个字符。