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

This is a discussion on keeping the printf() in the same place on the screen within the C Programming forums, part of the General Programming Boards category; I am new to C programming. I have looked through the tutorials and have not been able to find this. ...

  1. #1
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    keeping the printf() in the same place on the screen

    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

  2. #2
    Registered User claudiu's Avatar
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    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?

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    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:
    Code:
    while ( ... )
    {
       ...
       printf("\rCount: %3d", countVar++);
    }
    or I printed backspaces:
    Code:
    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.

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    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():

    Code:
    #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;
    }

  5. #5
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    Thanks

    Thank for the help!

  6. #6
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    ANSI escape sequences are portable and should work on all terminals.
    Code:
    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");
    }

  7. #7
    Guest Sebastiani's Avatar
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    Quote Originally Posted by itCbitC View Post
    ANSI escape sequences are portable and should work on all terminals.
    Doesn't seem to work on Windows7, though.
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

  8. #8
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    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.

  9. #9
    Guest Sebastiani's Avatar
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    Quote Originally Posted by Adak View Post
    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.
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

  10. #10
    spurious conceit MK27's Avatar
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    Lightbulb "\b" is for backspace

    There is also the easy way:
    Code:
    #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;
    }
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  11. #11
    Guest Sebastiani's Avatar
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    Quote Originally Posted by MK27 View Post
    There is also the easy way:
    Code:
    #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.
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

  12. #12
    spurious conceit MK27's Avatar
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    Quote Originally Posted by Sebastiani View Post
    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???
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  13. #13
    Guest Sebastiani's Avatar
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    Quote Originally Posted by MK27 View Post
    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).
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

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