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[34]non-buffering stdin reading

   [35]Ask Question
   Asked 13 years, 1 month ago
   Modified [36]9 years ago
   Viewed 6k times
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   My test application is
#include <sys/types.h>
#include <sys/wait.h>
#include <signal.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <fcntl.h>

int main(int argc, char *argv[], char *envp[]) {
  int fd[2];

  if(pipe(fd) < 0) {
    printf("Can\'t create pipe\n");
    exit(-1);
  }

  pid_t fpid = fork();
  if (fpid == 0) {
    close(0);
    close(fd[1]);
    char *s = (char *) malloc(sizeof(char));
    while(1) if (read(fd[0], s, 1)) printf("%i\n", *s);
  }
  close(fd[0]);
  char *c = (char *) malloc(sizeof(char));
  while (1) {
    if (read(0, c, 1) > 0) write(fd[1], c, 1);
  }
  return 0;
}

   I want to see char-code after each entered char. But in fact *s is
   printed only after '\n' in the console. So seems like stdin (file with
   desc 0) is buffered. But the read function is buffer-less, isn't it?
   Where am I wrong.

   UPD: I use linux.

   So the solution is
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <termios.h>

int main(int argc, char *argv[], char *envp[]) {
  int fd[2];

  if(pipe(fd) < 0) {
    printf("Can\'t create pipe\n");
    exit(-1);
  }

  struct termios term, term_orig;

  if(tcgetattr(0, &term_orig)) {
    printf("tcgetattr failed\n");
    exit(-1);
  }

  term = term_orig;

  term.c_lflag &= ~ICANON;
  term.c_lflag |= ECHO;
  term.c_cc[VMIN] = 0;
  term.c_cc[VTIME] = 0;

  if (tcsetattr(0, TCSANOW, &term)) {
    printf("tcsetattr failed\n");
    exit(-1);
  }

  pid_t fpid = fork();
  if (fpid == 0) {
    close(0);
    close(fd[1]);
    char *s = (char *) malloc(sizeof(char));
    while(1) if (read(fd[0], s, 1)) printf("%i\n", *s);
  }
  close(fd[0]);
  char *c = (char *) malloc(sizeof(char));
  while (1) {
    if (read(0, c, 1) > 0) write(fd[1], c, 1);
  }
  return 0;
}

     * [37]c
     * [38]linux
     * [39]unix
     * [40]posix

   [41]Share
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   [42]edited Dec 1, 2010 at 19:10
   Ximik
   asked Dec 1, 2010 at 18:54
   Ximik's user avatar
   [43]XimikXimik
   2,45533 gold badges2727 silver badges5353 bronze badges
   2
     * 1
       Note that this has nothing to do with buffering.
       - [44]Simon Tóth
       Dec 1, 2010 at 19:05
     * Shouldn't the code reset the terminal attributes in the parent to
       term_orig before exiting? You should probably also have the child
       exit at some point -- it will continually get 0 from read() after
       the parent shuts up shop. However, the parent is also in an
       infinite loop; the processes only end when signalled. You really
       need a signal handler that calls tcsetattr() with the original
       terminal values for the main signals you're likely to get (that can
       be handled): HUP, INT, QUIT perhaps, PIPE and TERM is a good set.
       You can't do anything about KILL or STOP, of course.
       - [45]Jonathan Leffler
       Jan 19, 2021 at 3:27

   [46]Add a comment  |

3 Answers 3

   Sorted by: [47]Reset to default
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   Unfortunately, the behavior you're looking for is not possible with
   standard ANSI C, and the default mode for UNIX terminal I/O is
   line-oriented, which means you will always need an inputted \n
   character to retrieve the input. You'll need to use terminal I/O
   facilities that let you program in [48]non-canonical mode, so that each
   key-press triggers an event. On Linux/UNIX, you can look into the
   <termios.h> header, or the [49]ncurses library.
   [50]Share
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   [51]edited May 23, 2017 at 10:29
   Community's user avatar
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   111 silver badge
   answered Dec 1, 2010 at 18:56
   Charles Salvia's user avatar
   [53]Charles SalviaCharles Salvia
   52.6k1414 gold badges130130 silver badges142142 bronze badges
   1
     * 4
       @Ximik, yes and they are not using standard ANSI C. Most use
       external libraries such as ncurses or termcap.
       - [54]Charles Salvia
       Dec 1, 2010 at 18:59

   [55]Add a comment  |
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   5
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   It seems to me that your solution is a little bit complicated. Still
   don't understand why do you need pipe and 2 process.
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <termios.h>

int main(int argc, char *argv[], char *envp[]) {
  struct termios term, term_orig;

  if(tcgetattr(0, &term_orig)) {
    printf("tcgetattr failed\n");
    exit(-1);
  }

  term = term_orig;

  term.c_lflag &= ~ICANON;
  term.c_lflag |= ECHO;
  term.c_cc[VMIN] = 0;
  term.c_cc[VTIME] = 0;

  if (tcsetattr(0, TCSANOW, &term)) {
    printf("tcsetattr failed\n");
    exit(-1);
  }

  char ch;
  while (1) {
    if (read(0, &ch, 1) > 0)
      printf(" %d\n", ch);
  }
  return 0;
}

   [56]Share
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   [57]edited Dec 30, 2014 at 12:56
   answered Dec 30, 2014 at 12:14
   Sam Toliman's user avatar
   [58]Sam TolimanSam Toliman
   12311 silver badge55 bronze badges

   [59]Add a comment  |
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   3
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   Unix buffers your tty characters inside the kernel in part so that
   programs don't have to individually handle line editing unless they
   want to.

   You can instruct the tty driver to give you the bytes immediately.
   There are various libraries that make this a bit easier than using the
   raw ioctl. You might start with termios(3).
   [60]Share
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   answered Dec 1, 2010 at 18:59
   DigitalRoss's user avatar
   [61]DigitalRossDigitalRoss
   144k2525 gold badges250250 silver badges330330 bronze badges
   1
     * 3
       And sadly nobody got around to improving this kernel-level line
       editing to make it actually usable... In principle it could be
       nearly as nice as readline.
       - [62]R.. GitHub STOP HELPING ICE
       Dec 1, 2010 at 20:46

   [63]Add a comment  |

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