[Courses] [C] format specifiers (was: yet one more question =))

[Katie Bechtold]

On Sat, Nov 09, 2002 at 12:00:39AM -0500, Morgon Kanter wrote:
> What is up with this formatting? :
> printf("%6d", blah);
>  - or -
> printf("%6.2d", blah);

The wonderful world of format specifiers!  If you specify an integer
right after the percent sign, that is the minimum field width.  For
example, while

printf("%d\n", 82);
produces:
82
 
printf("%6d\n", 82);
produces:
    82

If you lead the integer after the percent sign with a 0, it pads the
left side of the output with 0's.

If you specify an integer after a decimal point after the percent
sign, as in %.2f, that's the number of digits after the decimal
point that are displayed.  This has no effect on integers; looks
like it's only useful for floating point numbers, which are normally
displayed with a certain number of digits following the decimal
point. 

Here's an experimental program:

int main(int argc, char *argv[]) {

        int a = 82;
        float b = 1.23;

        printf("%d\n", a);
        printf("%6d\n", a);
        printf("%06d\n", a);
        printf("%6.2d\n", a);

        printf("%f\n", b);
        printf("%6f\n", b);
        printf("%06f\n", b);
        printf("%.2f\n", b);
        printf("%3f\n", b);

        return 0;
}

And here's its output:

82
    82
000082
    82
1.230000
1.230000
1.230000
1.23
1.230000
  

> also, weird, you don't need to include stdio for printf to work.

I think that's weird too.  When I look at printf's man page and see
#include <stdio.h> included in the SYNOPSIS, I kind of assume that's
where printf's defined and that I'd need to include it in my program
that calls printf. 

-- 
Katie Bechtold
http://katie-and-rob.org/