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Word access, high and low bytes

patroclus02 - May 9, 2006

		patroclus02	May 9, 2006
		If you do a move.W #$FF00, $200000, both /LWR and /UWR will assert, and 16 data bits will be driven. Low byte goes to $200000, and high byte goes to $200001 Ok, now, suppose you do move.W #$FF00, $200001 What happends there?? Low byte goes to $200001, and high byte goes to $200002? but how would /LWR and /UWR? Assembler warnings, that the adress is not aligned. But, what would happend?? Ok, this question is because I'm confused. Many ROMs connect A0 to 68k A1, ignoring A0. If you write 16 bits to a EEPROM, and you do move.W #$FF00, $200001 it will address same EEPROM address as move.W #$FF00, $200000 copied data would be the same? what happends if move.L #$FF00DDAA, $200000 it will write DDAA to $200000 and FF00 to $200002 ? :huh

		Mask of Destiny	May 9, 2006
		Originally posted by patroclus02@Tue, 2006-05-09 @ 02:18 PM Ok, now, suppose you do move.W #$FF00, $200001 What happends there?? A bus error. It's illegal to do word-wide access on an odd address on the 68K.

		cgfm2	May 9, 2006
		BTW, if you haven't read it I strongly suggest reading this manual: http://www.freescale.com/files/32bit/doc/r...l/MC6... It explains all the signals, all the bus cycles, and has pretty much everything you'd want to know about interfacing stuff to the 68000.

		patroclus02	May 9, 2006
		Thank you

		ExCyber	May 9, 2006
		Many ROMs connect A0 to 68k A1, ignoring A0. A0 isn't being ignored, it's not there in the first place because it's a 16-bit bus.