Re: bug in seq command?

Andrew D Jewell Sun, 23 Jan 2005 09:38:17 -0800

Please clear up my misunderstanding, because it seems to me that the fix is indeed trivial.

Note 1) The million below should, of course, be a parameter Note 2) I am not allowed to submit patches, so the below must be taken only as a suggestion.


*** seq.c~      Wed Dec  1 23:45:17 2004
--- seq.c       Sun Jan 23 07:55:07 2005
***************
*** 182,192 ****
  print_numbers (const char *fmt)
  {
    double i;

    for (i = 0; /* empty */; i++)
      {
        double x = first + i * step;
!       if (step < 0 ? x < last : last < x)
        break;
        if (i)
        fputs (separator, stdout);
--- 182,194 ----
  print_numbers (const char *fmt)
  {
    double i;
+   double delta = step/1000000.0;
+   double nlast = last + delta;

    for (i = 0; /* empty */; i++)
      {
        double x = first + i * step;
!       if (step < 0 ? x < nlast : nlast < x)
        break;
        if (i)
        fputs (separator, stdout);

At 10:15 PM -0800 1/21/05, Paul Eggert wrote:

"Luis A. Florit" <[EMAIL PROTECTED]> writes:

> the output of the command: seq -w 1 0.2 4

 is the sequence

        1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8

Why there is no 4.0 in the last sequence?


Floating point roundoff error.  It's documented in the manual.

If you can submit a fix, please do!  It would improve seq.
(Warning: it won't be trivial.)


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Re: bug in seq command?

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