偷看输入缓冲区，并刷新用C额外的字符 [英] peek at input buffer, and flush extra characters in C

问题描述

如果我想获得在C一个字符输入，我将如何检查，看看是否额外的字符被送到，如果是的话，我会怎么清除？

If I want to receive a one character input in C, how would I check to see if extra characters were sent, and if so, how would I clear that?

有没有这就像GETC（标准输入）功能，但不提示用户输入一个字符，这样我就可以只把而（GETC（标准输入）= EOF！）; ？或者一个函数来在缓冲区中的下一个字符偷看，如果不返回NULL（或任何会在那里），我可以打电话（另）一个函数，它刷新标准输入？

Is there a function which acts like getc(stdin), but which doesn't prompt the user to enter a character, so I can just put while(getc(stdin)!=EOF);? Or a function to peek at the next character in the buffer, and if it doesn't return NULL (or whatever would be there), I could call a(nother) function which flushes stdin?

所以，现在和scanf函数，似乎是在做的伎俩，但有没有办法得到它读全的字符串，直到换行？而不是最近的空白？我知道我可以只是把％s％s％S或任何入格式字符串，但我能处理的空间任意号码？

So right now, scanf seems to be doing the trick but is there a way to get it to read the whole string, up until the newline? Rather than to the nearest whitespace? I know I can just put "%s %s %s" or whatever into the format string but can I handle an arbitrary number of spaces?

推荐答案

您不能刷新输入流。如果你这样做，你会被调用未定义的行为。最好的办法是做：

You cannot flush the input stream. You will be invoking undefined behavior if you do. Your best bet is to do:

int main() {
  int c = getchar();
  while (getchar() != EOF);
  return 0;
}

要使用 scanf函数魔法：

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

#define str(s) #s
#define xstr(s) str(s)
#define BUFSZ 256

int main() {
  char buf[ BUFSZ + 1 ];
  int rc = scanf("%" xstr(BUFSZ) "[^\n]%*[^\n]", buf);
  if (!feof(stdin)) {
    getchar();
  }
  while (rc == 1) {
    printf("Your string is: %s\n", array);
    fflush(stdout);
    rc = scanf("%" xstr(LENGTH) "[^\n]%*[^\n]", array);
    if (!feof(stdin)) {
        getchar();
    }
   }
   return 0;
}

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