Path: utzoo!censor!geac!torsqnt!tmsoft!list-injector
Newsgroups: can.usrgroup
Subject: C code
Message-ID: <9003010101.AA08961@redvax>
Date: 1 Mar 90 01:01:35 GMT
Reply-To: redvax!hugh (D. Hugh Redelmeier)
Distribution: ont
Lines: 60

| From: Steve Bird <steve@cohort.UUCP>
|
|  Here's another interesting piece. 
|
|      /* floaterr.c--demonstrates round-off error */
|      # include <stdio.h>
|      main()
|        {
|          float a,b;
|          b = 2.0e20 + 1.0;
|          a = b - 2.0e20;
|          printf("%f \n",a);
|        }
|
|   When compiled the program returns the number 4008175468544.000000 .
|   Now when the program is modified to read :
|
|      /* floaterr.c--demonstrates round-off error */
|      # include <stdio.h>
|      main()
|        {
|          float a,b;
|          b = 2.0e20 + 1.0;
|          a = 2.0e20;
|          printf("%f \n",b - a);
|        }
|
|   The program returns 0.000000 . Why ?

In one case, you are evaluating
	((double) (float) 2.0e20) - 2.0e20
In the other case, you are evaluating
	((double) (float) 2.0e20) - ((double) (float) 2.0e20)

The former represents the roundoff in representing 2.0e20 in single
precision.  The latter ought to be zero.  Here is a simpler version
of your programs.  Note that the "+ 1.0" has no effect.

  /* floaterr.c--demonstrates round-off error */
  # include <stdio.h>
  int
  main()
    {
      printf("%f\n",((double) (float) 2.0e20) - 2.0e20);
        /*4008175468544.000000*/

      printf("%f\n", ((double) (float) 2.0e20) - ((double) (float) 2.0e20));
        /*0.000000*/

      return 0;
    }

For questions like this, it is useful to know the compiler and
machine.  It appears to use IEEE f.p. so I was able the reproduce it
using RedC on my Sun 3.

Hugh Redelmeier
{utcsri, yunexus, uunet!attcan, utzoo, hcr}!redvax!hugh
When all else fails: hugh@csri.toronto.edu
+1 416 482-8253