Message-ID: <3ADE8970.6366CCAE@earthlink.net>
From: Martin Ambuhl <mambuhl AT earthlink DOT net>
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Subject: Re: Atof
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Date: Wed, 18 Apr 2001 18:43:59 GMT
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To: djgpp AT delorie DOT com
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Igor Bujna wrote:
> 
> Hi,
> i have this problem with atof();
> 
> char *buf = "-21.345\0";
> float f = 0;
> 
> f = atof ( buf );
> printf("%f",f);
> 
> , then i have this "-21.344999".
> What i must to do for it will be OK.

You don't have a problem with atof().  You have a problem with trying
to get 8 digits of significance from a type only guaranteed 6 digits of
significance:

#include <stdlib.h>
#include <stdio.h>
#include <float.h>

int main(void)
{
    const char *buf = "-21.345\0";
    float f = 0;
    double d = 0;

    f = atof(buf);
    printf("unadorned %%f applied to float: %f\n", f);
    printf("%%.3f applied to float: %.3f\n", f);
    printf("%%.*f (for FLT_DIG-2) applied to float: %.*f\n\n",
           FLT_DIG - 2, f);

    d = atof(buf);
    printf("unadorned %%f applied to double: %f\n", d);
    printf("%%.3f applied to double: %.3f\n", d);
    printf("%%.*f (for DBL_DIG-2) applied to double: %.*f\n\n",
           DBL_DIG - 2, d);

    return 0;
}

unadorned %f applied to float: -21.344999
%.3f applied to float: -21.345
%.*f (for FLT_DIG-2) applied to float: -21.3450

unadorned %f applied to double: -21.345000
%.3f applied to double: -21.345
%.*f (for DBL_DIG-2) applied to double: -21.3450000000000