scanf [^ \ n]循环执行，除了第一次以外，不停止等待用户输入 [英] scanf[^\n] in a loop , not stop to wait for user input except for 1st time

问题描述

我需要使用scanf(％[^''\ n"]输入一个句子.输入是循环的，我还有其他代码块来处理它.

如果我仅使用^ \ n以外的scanf(％s)，则每个时间循环中，控制台将停止等待用户输入，这是我需要的正确例程.

但是，如果我使用scanf(％[^ \ n]或％[^''\ n'']，仅在控制台第一次等待用户输入时，它将继续连续显示printf("Input:")而不会停止等待新输入.

我尝试将数组分配为null，可用数组空间，刷新流，但无济于事.

简单的示例代码和示例错误的输出:

I need to use scanf("%[^''\n'']" to input a sentence. And the input is in a loop, I have other code block to process it.

If I just use scanf(%s) other than ^\n,each time loop the console will stop to wait user input, that''s correct routine I need.

However, if I use scanf(%[^\n] or %[^''\n''], only the 1st time the console wait for user input, then it will continue display printf("Input:") continuously without stop to wait the new input.

I tried assign array to null, free array space, flush stream, not helpful.

Simple sample code and sample incorrect output:

#include <stdio.h>

void userInput(){

  char* in;
  in=malloc(sizeof(char)*20);
  printf("Input: ");
  scanf("%[^'\n']",in);
  fflush(stdin);
  free(in);
}

void main(void) {

  for(;;)
    userInput();
}</stdio.h>

样本故障输出:

Sample Malfunction output:

Input:I enter something
Input: Input: Input:........Input

推荐答案

您的scanf()语句错误，它不使用正则表达式.一种更简单的方法是使用 gets() [

Your scanf() statement is wrong, it does not use regular expressions. An easier way would be to use gets()[^]. There is also little point in using malloc() and free() in your subroutine. Just allocate the required number of characters on the stack thus:

char in[20];

尽管如果您想输入长字符串，则20可能还不够.

Although 20 may not be enough if you want to enter a long string.

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