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Path: utzoo!watmath!clyde!rutgers!ames!amdahl!drivax!alexande
From: alexande@drivax.UUCP
Newsgroups: comp.arch,comp.sys.m68k
Subject: Bit operations on 68K
Message-ID: <816@drivax.UUCP>
Date: Thu, 22-Jan-87 12:38:24 EST
Article-I.D.: drivax.816
Posted: Thu Jan 22 12:38:24 1987
Date-Received: Mon, 26-Jan-87 01:48:04 EST
References: <760@orcisi.UUCP> <112@lmi-angel.UUCP>
Reply-To: alexande@drivax.UUCP (Mark Alexander)
Organization: Kludges 'R' Us, Inc.
Lines: 29
Xref: watmath comp.arch:240 comp.sys.m68k:169

In article <112@lmi-angel.UUCP> wsr@lmi-angel.UUCP (Wolfgang Rupprecht) writes:
>Look at the BSET, BCLR, BTST (bit set, clear and test) instructions.
>If you do a:
>
>	BSET.L d0, somelocation
>
>with #31 in d0, you have set the MSB (byte 0!) of the long-word.

This is not completely correct.  There is no BSET.L instruction, only BSET.
It works on a destination that is either a 32-bit register, or a byte from
memory. Here is what the Motorola book says about BSET:

	If a data register is the destination, then the bit numbering is
	modulo 32, allowing bit manipulation on all bits in a data register.
	If a memory location is the destination, a byte is read
	from that location, the bit operation is performed using
	the bit number modulo 8, and the byte written back to
	the location with zero referring to the least-significant bit.

I didn't take the manual's word for it.  I tried it on a Motorola VME/10.

>This is *LOW* endian.

Yes, as far as bit numbering goes.  No, as far as byte numbering goes.
That's the inconsistency in the 68K architecture.
-- 
Mark Alexander	...{hplabs,ucbvax!decvax}!decwrl!pyramid!amdahl!drivax!alexande
"This then is my story.  I have reread it.  It has bits of marrow
sticking to it, and blood, and beautiful bright-green flies."  --Nabokov