Mail Archives: djgpp/2000/08/01/12:30:29
| From: | Radical NetSurfer <radsmail AT juno DOT com>
|
| Newsgroups: | comp.os.msdos.djgpp
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| Subject: | Re: modf BROKEN ??
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| Date: | Tue, 01 Aug 2000 12:23:06 -0400
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| Message-ID: | <f8udosk8samrov7j0156qkuk31aga0essa@4ax.com>
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| References: | <crdcosk167oh81vpnjfqeg1bnls5qqg9ld AT 4ax DOT com>
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| To: | djgpp AT delorie DOT com
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| DJ-Gateway: | from newsgroup comp.os.msdos.djgpp
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| Reply-To: | djgpp AT delorie DOT com
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At 5.0, 6.0, etc, Fract display as "1.0" because of a rounding error
in the printf output. Its only an "optical illusion", and if
%3.21 is used, you won't see it.
the question is obvious:
HOW do I maintain a %3.3f output, but NOT get the
1.0 "optical illusion" ???
Thanks!
On Mon, 31 Jul 2000 22:37:40 -0400, Radical NetSurfer
<radsmail AT juno DOT com> wrote:
>Why does the following program INCORRECTLY display
>the fractional part?
>
>#include <stdio.h>
>#include <stdlib.h>
>#include <conio.h>
>#include <math.h>
>
>#include <pc.h>
>#include <unistd.h>
>
>int main(int argc, char **argv) {
>double i, f;
>double v, y;
>
> printf("Observe the \"Rounded\" Value of Fint below:\n");
> printf("Use ESC to exit.\n\n");
>
> for(v=0.0; v< 10; v+=0.1) {
> f = modf(v, &i); // fractional = modf(argument, &integral);
> if ( f <0.5 ) y = floor(v);
> else y = ceil(v);
>
> printf("Value: %lf Fint: %lf Int: %lf Frac: %lf\n",
> v, y, i, f );
> if ( getch() == 27 ) break;
> }
>
>return (0);
>} //main
>------------------------------------------------------------------------------------------------------
>well?
>
>Email always welcomed: radsmail AT juno DOT com
>http://members.tripod.com/~RadSurfer/
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