[TUHS] PDP-11 DIV instruction lossage

[jnc@mercury.lcs.mit.edu (Noel Chiappa)]

So, I have discovered, to my astonishment, that the double-word version of the
DIV instruction on the PDP-11 won't do a divide if the result won't fit into
15 bits. OK, I can understand it bitching if the quotient wouldn't fit into 16
bits - but what's the problem with returning an unsigned quotient?

And, just for grins, the results left in the registers which hold the quotient
and remainer is different in the -11/23 (KDF11-A) and the -11/73 (KDJ11-A).
(Although, to be fair, the PDP-11 Architecture Manual says 'register contents
are unpredictable if there's an overflow'.)

Oh well, guess I'll have to redo my kludgy fix to gmtime() (the distributed
version of which in V6 qhas a problem when the number of 8-hour periods since
the epoch overflows 15 bits)! I guess I'll have to fix it for real, instead of
my kludgy fix (which extended it to work for 16-bit results). :-)


I discovered this when I plugged in an -11/73 to make sure the prototype QSIC
(our RK11/etc emulator for the QBUS) worked with the -11/73 as well as the
-11/23 (which is what we'd mostly been using - when we first started working
on the DMA and interrupts, we did try them both). I noticed that with the
-11/73, the date printed incorrectly:

  Sun Mar 10 93:71:92 EST 1991

After a certain amount of poking and prodding, I discovered the issue - and
on further reading, discovered the limitation to 15-bit results.


For those who are interested in the details, here's a little test program that
displays the problem:

       r = ldiv(a, b, d);
       m = ldivr;

       printf("a: 0%o %d.  b: 0%o %d.  d: 0%o %d.\n", a, a, b, b, d, d);
       printf("q: 0%o %d.  r: 0%o %d.\n", r, r, m, m);

and, for those who don't have V6 source at hand, here's ldiv():

        mov     2(sp),r0
        mov     4(sp),r1
        div     6(sp),r0
        mov     r1,_ldivr
        rts     pc

So here are the results, first from a simulator:

  tld 055256 0145510 070200
  a: 055256 23214.  b: 0145510 -13496.  d: 070200 28800.
  q: 0147132 -12710.  r: 037110 15944.

This is _mathematically_ correct: 055256,0145510 = 1521404744., 070200 =
28800., and 1521404744./28800. = 0147132.

And on the -11/23:

  a: 055256 23214.  b: 0145510 -13496.  d: 070200 28800.
  q: 055256 23214.  r: 037110 15944.

Note that the returned 'quotient' is simply the high part of the dividend.

And on the -11/73:

  a: 055256 23214.  b: 0145510 -13496.  d: 070200 28800.
  q: 055256 23214.  r: 037110 15944.

Note that in addition to the quotient behaviour, as with the /23, the
'remainder' is the low part of the dividend.

	    Noel