如何检测在Linux下C GUI程序键presses不提示用户？ [英] How to detect key presses in a Linux C GUI program without prompting the user?

问题描述

如何检测在C键盘事件而不提示在linux用户？这就是程序运行应当由pressing任意键终止。
任何人都可以请帮助呢？

how to detect a keyboard event in C without prompting the user in linux? That is the program running should terminate by pressing any key. can anyone please help with this?

推荐答案

您必须修改使用的termios终端设置。见史蒂文斯及放大器;拉戈第二版UNIX环境高级编程这解释了为什么tcsetattr（）可以在无需将所有终端characteristcs successfuly回来，为什么你看到什么看起来是多余的调用tcsetattr（）。

You have to modify terminal settings using termios. See Stevens & Rago 2nd Ed 'Advanced Programming in the UNIX Environment' it explains why tcsetattr() can return successfuly without having set all terminal characteristcs, and why you see what looks to be redundant calls to tcsetattr().

这是在UNIX ANSI C：

This is ANSI C in UNIX:

#include <sys/types.h>
#include <sys/time.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <unistd.h>
#include <termios.h>
#include <errno.h>

int checktty(struct termios *p, int term_fd)
{
    struct termios ck;
    return (
       tcgetattr(term_fd, &ck) == 0 &&
      (p->c_lflag == ck.c_lflag) &&
      (p->c_cc[VMIN] == ck.c_cc[VMIN]) &&
      (p->c_cc[VTIME] == ck.c_cc[VMIN])
    );
}


int
keypress(int term_fd)
{
     unsigned char ch;
   int retval=read(term_fd, &ch, sizeof ch);
   return retval;
}

int   /* TCSAFLUSH acts like fflush for stdin */
flush_term(int term_fd, struct termios *p)
{
   struct termios newterm;
   errno=0;
   tcgetattr(term_fd, p);  /* get current stty settings*/

   newterm = *p; 
   newterm.c_lflag &= ~(ECHO | ICANON); 
   newterm.c_cc[VMIN] = 0; 
   newterm.c_cc[VTIME] = 0; 

   return( 
       tcgetattr(term_fd, p) == 0 &&
       tcsetattr(term_fd, TCSAFLUSH, &newterm) == 0 &&
       checktty(&newterm, term_fd) != 0
   );
}
void 
term_error(void)
{
     fprintf(stderr, "unable to set terminal characteristics\n");
     perror("");                                                
     exit(1);                                                   
}


void
wait_and_exit(void)
{
    struct timespec tsp={0,500};  /* sleep 500 usec (or likely more ) */
    struct termios  attr;
    struct termios *p=&attr;
    int keepon=0;
    int term_fd=fileno(stdin);

    fprintf(stdout, "press any key to continue:");
    fflush(stdout);
    if(!flush_term(term_fd, p) )
       term_error();
    for(keepon=1; keepon;)
    {
        nanosleep(&tsp, NULL);
        switch(keypress(term_fd) )
        {
              case 0:
              default:
                 break;
            case -1:
                 fprintf(stdout, "Read error %s", strerror(errno));
                 exit(1);
                 break;
            case 1:       /* time to quit */
                 keepon=0;
                 fprintf(stdout, "\n");
                 break;                 
        } 
    }
    if( tcsetattr(term_fd, TCSADRAIN, p) == -1 && 
          tcsetattr(term_fd, TCSADRAIN, p) == -1 )
          term_error();
    exit(0);
}

int main()
{
      wait_and_exit();
      return 0;  /* never reached */
}

在了nanosleep调用那里prevent从吞噬系统资源$ C ​​$ C。你可以叫尼斯（），而不是使用了nanosleep（）。所有这一切确实是坐等一个按键，然后退出。

The nanosleep call is there to prevent the code from gobbling up system resources. You could call nice() and not use nanosleep(). All this does is sit and wait for a keystroke, then exit.

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