🚀 go-pugleaf

In article <2804@bobkat.UUCP>, m5@bobkat.UUCP (Mike McNally ) writes:
> How do clever iAPX 386 users (or iAPX *8[68] users for that matter) 
> implement C-style string copies?  Seems to me that there are two
> approaches:
> 
> loop:   lodsb
>         stosb
>         cmp     AL, 0
>         loopne  loop
> 
> The other approach would be to SCASB to the end of the source string,
> then move that many bytes with REP MOVSB.  Neither way is particularly
> appealing.  So like, how do the pros do it?  What's the deal?

Here's a quick comparison of the two approaches above, using
8086 cycle counts.  I've left out the overhead of setting up
all the appropriate registers before getting to the loop itself.

Repeat_Here:  LODSW                ; 12 cycles
	      OR      AL,AL        ; 3 cycles
	      JZ      Exit         ; 4 cycles to stay in the loop
	      STOSW                ; 11 cycles
	      OR      AH,AH        ; 3 cycles
	      LOOPNZ  Repeat_Here  ; 19 cycles to stay in the loop
Exit:                              ; Total: 52 cycles per 2 characters

	      REP     SCASB        ; 15 cycles per character
	      REP     MOVSW        ; 17 cycles per 2 characters
				   ; Total: 47 cycles per 2 characters

But why try to be so clever?  My approach is MUCH simpler: just
move a number of characters equal to the lesser of the declared
lengths of the two strings.  Since REP MOVSW is only 17 cycles
per 2 characters, this is more efficient whenever the shorter of
the strings is longer than 1/3 its declared maximum length.  On
the average this wins 2 out of 3 times, and the code is both
smaller and easier to read.
-- 
{hplabs,amdahl,ames}!bnrmtv!perkins         --Henry Perkins

It is better never to have been born.  But who among us has such luck?
One in a million, perhaps.