keeping the printf() in the same place on the screen

[jm0825]

I am new to C programming. I have looked through the tutorials and have not been able to find this.

I have not been able to find a way to process a printf statement and keep it it the same place on the screen.

I have a calculation that repeats its self in a loop until it has completed its loop.
ie; count from 1 to 100

I want to display printf("The current count is:",variable for count);

I want to display the calculation as it loops to stay in the same spot on the screen instead of printing a list that goes down the monitor.

Is there any way I can do this?

Jack

[claudiu]

You could add an instruction to clear the screen after each print I guess. Not really sure you would actually see anything though. Why would you want to do that anyway?

[arpsmack]

I was able to do something like this before, but I can't remember exactly how.

I believe I either printed a carriage return without a linefeed before the text each time I printed it:

while ( ... )
{
   ...
   printf("\rCount: %3d", countVar++);
}

or I printed backspaces:

printf("Count: ");

while ( ... )
{
   ...
   printf("%3d\b\b\b", countVar++);
}

Whatever it was, I'm sure it's not portable at all. If you have a non-standard console library available (conio.h for Windows as an example), you can use caret positioning to achieve the same thing.

[Adak]

There are 3 ways to do it. One uses the old conio.h include file header, or ncurses library.

Second way is to use the Windows API SetConconsoleCursorPosition, an example of which is in our FAQ under "How do I clear the screen", advanced.

This is the all C way, using the escape char \b and printf():

#include <stdio.h>
#include <dos.h>      //just for the delay() function


int main() {
  int i, n ; 
  printf("\n\n The Number is: ");
  for(i = 50; i > -1; i--)  {
    printf("%2d", i);
    delay(400);
    printf("\b\b");
  }


  printf("\n\n\t\t\t     press enter when ready");

  i = getchar();
  return 0;
}

[jm0825]

Thank for the help!

[itCbitC]

ANSI escape sequences are portable and should work on all terminals.

printf("\033[2J");        /*  clear the screen  */
printf("\033[H");         /*  position cursor at top-left corner */

for (i=1; i<=100; i++) ) {
    printf("The current count is: %d", i);
    fflush(stdout);
    sleep(1);
    printf(i < 100 ? "\033[H" : "\n");
}

[Sebastiani]

ANSI escape sequences are portable and should work on all terminals.

Doesn't seem to work on Windows7, though.

[Adak]

ANSI escape sequences are only going to work if your system uses the ansi.sys driver.

I have WinXP and ANSI escape sequences will not work on this system.

[Sebastiani]

ANSI escape sequences are only going to work if your system uses the ansi.sys driver.

I have WinXP and ANSI escape sequences will not work on this system.

Ah, okay. Damn Windows.

[MK27]

There is also the easy way:

#include <stdio.h>

int main() {
	int i = 0;

	printf("Count is: %2d", i);
	for (;i<60;i++) {
		printf("\b\b%2d",i); fflush(stdout);
		sleep(1);
	}

	return 0;
}

[Sebastiani]

There is also the easy way:

#include <stdio.h>

int main() {
    int i = 0;

    printf("Count is: %2d", i);
    for (;i<60;i++) {
        printf("\b\b%2d",i); fflush(stdout);
        sleep(1);
    }

    return 0;
}

I guess. It is a bit of a kludge, though.

[MK27]

I guess. It is a bit of a kludge, though.

Yes, better to use ANSI sequences or import an entire console library What have YOU been smoking???

[Sebastiani]

What have YOU been smoking???

I won't say, but I can assure you that I haven't smoked enough of it...yet (it's still early).