NaN and equal

[James Clark <jjc@jclark.com>]

With Emacs 21.3.1, i686-pc-linux-gnu

  (equal 0.0e+NaN 0.0e+NaN)

returns nil.  This does not seem consistent with the documentation of 
equal.  I would suggest that (= 0.0e+NaN 0.0e+NaN) continues to return 
nil, for consistency with the IEEE floating point standard (which of 
course says that NaN != NaN), but that (equal 0.0e+NaN 0.0e+NaN) return 
t, so that hash-tables and alists with NaN can work.

Here's a patch.  I don't think this should cause any problems on systems 
without NaN.

*** /home/jjc/var/build/emacs-21.3/src/fns.c~	2002-12-04 
19:38:32.000000000 +0700
--- /home/jjc/var/build/emacs-21.3/src/fns.c	2003-08-12 
23:04:52.000000000 +0700
***************
*** 1976,1982 ****
     switch (XTYPE (o1))
       {
       case Lisp_Float:
!       return (extract_float (o1) == extract_float (o2));

       case Lisp_Cons:
         if (!internal_equal (XCAR (o1), XCAR (o2), depth + 1))
--- 1976,1990 ----
     switch (XTYPE (o1))
       {
       case Lisp_Float:
!       {
! 	double d1, d2;
! 	
! 	d1 = extract_float (o1);
! 	d2 = extract_float (o2);
! 	/* If d is NaN, then d != d. Two NaNs should be equal
! 	   even though they are not =. */
! 	return d1 == d2 || (d1 != d1 && d2 != d2);
!       }

       case Lisp_Cons:
         if (!internal_equal (XCAR (o1), XCAR (o2), depth + 1))


James