how to get position of cursur in C++

rahulvishwakarma · 02-19-2020, 08:23 AM

hi to all, I've centos 7.5 in VM, and i am using C::B, now i want to know position of cursur in terminal. how to do that. i googled and find below soution but i am not getting its point. code is:-

Code:

#include <errno.h>
#include <fcntl.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>

int main (void)

{
  int ttyfd = open ("/dev/tty", O_RDWR);
  if (ttyfd < 0)
    {
      printf ("Cannot open /devv/tty: errno = %d, %s\r\n",
        errno, strerror (errno));
      exit (EXIT_FAILURE);
    }

  write (ttyfd, "\x1B[6n\n", 5);

  unsigned char answer[16];
  size_t answerlen = 0;
  while (answerlen < sizeof (answer) - 1 &&
         read (ttyfd, answer + answerlen, 1) == 1)
    if (answer [answerlen ++] == 'R') break;
  answer [answerlen] = '\0';

  printf ("Answerback = \"");
  for (size_t i = 0; i < answerlen; ++ i)
    if (answer [i] < ' ' || '~' < answer [i])
      printf ("\\x%02X", (unsigned char) answer [i]);
    else
      printf ("%c", answer [i]);
  printf ("\"\r\n");

  return EXIT_SUCCESS;
}

output is :-

Code:

[rahul@centos7client Release]$ ./poscurr 

^[[28;1R
         Answerback = "\x1B[28;1R"
[rahul@centos7client Release]$

what it does mean. please any one can explain this.

TB0ne · 02-19-2020, 10:16 AM

Quote:

Originally Posted by rahulvishwakarma

hi to all, I've centos 7.5 in VM, and i am using C::B, now i want to know position of cursur in terminal. how to do that. i googled and find below soution but i am not getting its point. code is:-

Code:

#include <errno.h>
#include <fcntl.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>

int main (void)

{
  int ttyfd = open ("/dev/tty", O_RDWR);
  if (ttyfd < 0)
    {
      printf ("Cannot open /devv/tty: errno = %d, %s\r\n",
        errno, strerror (errno));
      exit (EXIT_FAILURE);
    }

  write (ttyfd, "\x1B[6n\n", 5);

  unsigned char answer[16];
  size_t answerlen = 0;
  while (answerlen < sizeof (answer) - 1 &&
         read (ttyfd, answer + answerlen, 1) == 1)
    if (answer [answerlen ++] == 'R') break;
  answer [answerlen] = '\0';

  printf ("Answerback = \"");
  for (size_t i = 0; i < answerlen; ++ i)
    if (answer [i] < ' ' || '~' < answer [i])
      printf ("\\x%02X", (unsigned char) answer [i]);
    else
      printf ("%c", answer [i]);
  printf ("\"\r\n");

  return EXIT_SUCCESS;
}

output is :-

Code:

[rahul@centos7client Release]$ ./poscurr 

^[[28;1R
         Answerback = "\x1B[28;1R"
[rahul@centos7client Release]$

what it does mean. please any one can explain this.

Please see your other threads regarding very similar topics:
https://www.linuxquestions.org/quest...-a-4175669527/
https://www.linuxquestions.org/quest...ux-4175669529/

pan64 · 02-19-2020, 01:24 PM

which part is unclear?

berndbausch · 02-19-2020, 03:54 PM

I suggest you look up ANSI terminal escape sequences.

Code:

^[[28;1R

Row 28, column 1.

Edit: I don't know what's C::B. You program looks like regular C.

rahulvishwakarma · 03-15-2020, 09:58 AM

sorry for late, when i ran that program and it always shows "^[[Y;1R" where Y is row and x coordinate is 1 always. How to get rid of this.

TB0ne · 03-15-2020, 10:02 AM

Quote:

Originally Posted by rahulvishwakarma

sorry for late, when i ran that program and it always shows "^[[Y;1R" where Y is row and x coordinate is 1 always. How to get rid of this.

Sorry, but this sounds very much like homework. You have a task that has no real purpose, and you downloaded code that does what you want...and it now seems like you want us to rewrite/explain it all to you.

What have YOU figured out/thought about in regards to the code?? And chances are that code is returning exactly what it should...want a different number? Put the cursor somewhere else. And you do realize that the MOUSE cursor is far different than the TERMINAL cursor, right???

berndbausch · 03-15-2020, 10:53 AM

Quote:

Originally Posted by rahulvishwakarma

sorry for late, when i ran that program and it always shows "^[[Y;1R" where Y is row and x coordinate is 1 always. How to get rid of this.

This is the terminal's answer to the escape sequence "\x1B[6n\n". Do you not want to see this output? If so, perhaps there is a way to put the terminal in a mode that suppresses output on the screen.

EDIT: There is. Good old stty, how could I forget it. These days everything is graphical, but in the golden days, we had serial terminals when talking to UNIX (which was still better than punched cards, if you ask me):

Code:

stty -echo; ./yourprogram; stty echo

You do your own googling to find a suitable escape sequence or other programmatic way.