gpt4 book ai didi

c - 在 C 中安全丢弃可变长度的标准输入字符的正确方法是什么?

转载 作者:太空宇宙 更新时间:2023-11-04 08:28:16 25 4
gpt4 key购买 nike

我正在编写名为“Headfirst C”的 C 文本中的一些示例代码。我编写了一个演示信号处理的练习应用程序,并在完成本章后决定尝试一下。我是一名工程师,曾经在 LabVIEW(高度并发和直观的事件处理功能)中工作,所以我对使用警报和信号处理程序来生成周期性事件很感兴趣。我的问题是:

在以下示例代码中,丢弃位于标准输入中的可变数量的用户输入的正确方法或最佳实践是什么?我已经将这个小应用程序编写为演示,3 秒警报会触发恼人的“你好!”中断 fgets 调用的消息。然而,我注意到的是,如果用户在打字过程中被打断,当他最终按下回车键时,任何输入的文本(无论是否被打断)都会被回显。我想丢弃在用户按下回车键之前被打断的任何东西。

//Sample Program - Signal Handling & Alarms
//Header Includes
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#include <signal.h>

//Function Declarations
//Handler Functions
void diediedie(int sig);
void howdy(int sig);
//Handler Register Function
int catchSignal(int signum, void(*handler)(int));

//Variable declarations
//Declare interrupted flags
static int interrupted = 0;

//Program entrypoint
int main() {

//Register interrupt hander, catch errors
if(catchSignal(SIGINT, diediedie) == -1) {
fprintf(stderr, "Could not register interrupt handler");
exit(2);
}

//Register alarm handler, catch errors
if(catchSignal(SIGALRM, howdy) == -1) {
fprintf(stderr, "Could not register alarm handler");
exit(2);
}

//Create initial alarm trigger
alarm(3);

//Do something stupid while waiting for signals
char name[30];
printf("Enter your name: ");
//Keep waiting for user input even if interrupted by alarm signal
while(1) {
fgets(name, 30, stdin);
if(interrupted) {
// reset interrupted flag
interrupted = 0;
// ***** ADD CODE TO DISCARD INTERRUPTED INPUT HERE ******
}
else {
//echo user input and break out
printf("Hello %s\n", name);
break;
}
}

//end program
return 0;

}

//interrupt handler definition
void diediedie(int sig) {

//write some stuff, exit program
puts("Goodbye world!");
exit(1);

}

//alarm handler definition
void howdy(int sig) {
//set interrupted flag
interrupted = 1;
//write some annoying message
puts("howdy!");
//set another alarm trigger
alarm(3);
//**** COULD STDIN BE FLUSHED HERE? ****
}

//signal handler registration function definition
int catchSignal(int signum, void(*handler)(int)) {

//register handler
struct sigaction action;
action.sa_handler = handler;
sigemptyset(&action.sa_mask);
action.sa_flags = 0;
return sigaction(signum, &action, NULL);

}

在警报处理程序中是否适合执行此清除操作?请注意表明我对正确代码位置的想法的注释。

我考虑了以下几点:

while(getchar() != EOF) {}

我也想知道,当 fgets 等待用户输入并引发 SIGALRM 时会发生什么?该功能是否已终止?我观察到,如果我不包括 while 循环来检查中断标志并做出适当响应,程序将完成 fgets,在屏幕上转储一些垃圾(我假设 stdin 的当前状态?)并结束程序。

感谢您的任何建议!

最佳答案

在 Unix 中,信号处理程序与您的代码一起出现在带外。如果信号发生在阻塞系统调用的中间,系统调用将退出,errno 设置为 EINTR。但我相信 fgets() 正在为您处理此中断并继续进行而不会将控制权返回给您。

如果您使用的是基于 Unix 的操作系统并从命令行键入输入,那么这里真正发生的是您正在以熟化模式从终端读取数据。在按下回车之前,您的程序不会从 TTY 获取任何数据。您需要将终端设置为“原始”模式。以下是如何将其与您的代码集成的示例:

//Sample Program - Signal Handling & Alarms
//Header Includes
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#include <signal.h>
#include <termios.h>

//Function Declarations
//Handler Functions
void diediedie(int sig);
void howdy(int sig);
//Handler Register Function
int catchSignal(int signum, void(*handler)(int));

//Variable declarations
//Declare interrupted sa_flags
static int interrupted = 0;

static struct termios save_termios;

//Program entrypoint
int main() {

struct termios buf;
int fd = 1;

// This is derived from from Stevens, "Advanced Programming in the UNIX Environment"
if (tcgetattr(fd, &save_termios) < 0) /* get the original state */
return -1;

buf = save_termios;

buf.c_lflag &= ~(ECHO | ICANON | IEXTEN | ISIG);
/* echo off, canonical mode off, extended input
processing off, signal chars off */

buf.c_iflag |= BRKINT | ICRNL;
/* SIGINT on BREAK, CR-toNL on */

buf.c_cflag &= ~(CSIZE | PARENB);
/* clear size bits, parity checking off */

buf.c_cflag |= CS8;
/* set 8 bits/char */

buf.c_oflag &= ~(OPOST);
/* output processing off */

buf.c_cc[VMIN] = 1; /* 1 byte at a time */
buf.c_cc[VTIME] = 0; /* no timer on input */

if (tcsetattr(fd, TCSAFLUSH, &buf) < 0)
return -1;


//Register interrupt hander, catch errors
if(catchSignal(SIGINT, diediedie) == -1) {
fprintf(stderr, "Could not register interrupt handler");
exit(2);
}

//Register alarm handler, catch errors
if(catchSignal(SIGALRM, howdy) == -1) {
fprintf(stderr, "Could not register alarm handler");
exit(2);
}

//Create initial alarm trigger
alarm(3);

//Do something stupid while waiting for signals
char name[30];
printf("Enter your name: ");

//Keep waiting for user input even if interrupted by alarm signal
char nextchar = 0;
char *p;
p = name;
while(nextchar != '\n') {
nextchar = fgetc(stdin);
if (interrupted) {
// reset interrupted flag
interrupted = 0;

//Discard interrupted input by reseting 'p' to the start of the buffer
p = name;
*p = 0;
continue;
}
if (nextchar == '\n') {
*p = 0;
fputc('\r', stdout);
fputc('\n', stdout);
break;
}
// You'll have to handle some characters manually to emulate what the
// terminal does, or you could filter them out using a function like isprint()
//
if (nextchar == 127) {
// *** handle backspace
if (p > name) {
p--;
}
// TODO: To handle this right you'll have to backup the cursor on the screen
} else {
*p = nextchar;
p++;
}
fputc(nextchar, stdout);

// Handle buffer overflow
if (p-name == sizeof(name) - 1) {
*p = 0;
break;
}
}
// echo user input
printf("Input is: %s\r\n", name);
tcsetattr(1, TCSAFLUSH, &save_termios);
}

//interrupt handler definition
void diediedie(int sig) {

//write some stuff, exit program
puts("Goodbye world!");

tcsetattr(1, TCSAFLUSH, &save_termios);
exit(1);

}

//alarm handler definition
void howdy(int sig) {
//set interrupted flag
interrupted = 1;
//write some annoying message
puts("howdy!");
//set another alarm trigger
alarm(3);
}

// signal handler registration function definition
int catchSignal(int signum, void(*handler)(int)) {

//register handler
struct sigaction action;
action.sa_handler = handler;
sigemptyset(&action.sa_mask);
action.sa_flags = 0;
return sigaction(signum, &action, NULL);
}

注意需要保存原来的终端设置,在程序退出前恢复!如果遇到麻烦,您可能会破坏终端设置。从命令行使用 resetstty sane 来恢复正常的终端设置。有关 termios 数据结构的更多信息,请参阅手册页。

您还可以使用像 ncurses 这样的库处理原始输入。

关于c - 在 C 中安全丢弃可变长度的标准输入字符的正确方法是什么?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29451133/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com