Xref: utzoo comp.lang.c:11750 comp.std.c:249 sci.math:4360
Path: utzoo!attcan!uunet!husc6!mailrus!cornell!batcomputer!braner
From: braner@batcomputer.tn.cornell.edu (braner)
Newsgroups: comp.lang.c,comp.std.c,sci.math
Subject: Re: Floating point puzzle
Summary: Beware of conversion to double in passing to function
Keywords: floating point representation
Message-ID: <5810@batcomputer.tn.cornell.edu>
Date: 7 Aug 88 23:35:05 GMT
References: <3117@emory.uucp>
Reply-To: braner@tcgould.tn.cornell.edu (braner)
Organization: Cornell Theory Center, Cornell University, Ithaca NY
Lines: 33

In article <3117@emory.uucp> riddle@emory.uucp (Larry Riddle) writes:
>
>main()
>{
>	float x,y;
>	x = 1.0/10.0;
>	y = 1677721.0/16777216.0; 
>	printf("x: %x",x);
>	printf("%20.17f\n",x);
>	printf("y: %x",y);
>	printf("%20.17f\n",y);
>}
>
>Here is the output:
>
>x: 3fb99999 0.10000000149011612 
>y: 3fb99999 0.09999996423721313

Seems that the calculation of y is just different enough from 1/10 to cause
a difference in the last bit or so of the floats.  In the printf() calls the
floats are converted to doubles before being passed to the function, and
since the doubles have a larger exponent field the difference in the
mantissa is out of sight within the (long) int referred to in the "%x".
I suggest you try:

	float	x;
	long	*p;
	...
	p = (long *) &x;
	printf ("representation of x: %lx\n", *p);
	...and so on...

- Moshe Braner