From: Marko Mäkelä (msmakela_at_gmail.com)
Date: 2007-12-05 21:13:45
On Wed, Dec 05, 2007 at 05:16:30PM +0100, Antitrack@networld.at wrote: > > Maybe this will clear things: an harddisk > > sector contains 512 bytes. Using an 8 bits interface means that one can > > only access 256 of these 512 bytes. Writing $00, $01 .... $FF to a > > sector with my interface and reading the same sector using a PC will > > give $FF, $00, $FF, $01, $FF, $02 ..... $FF, $FE, $FF, $FF. > > Ah yes, this makes it much more clear for me. You cannot store more than 256 > bytes because you don't have a 9th line providing that extra bit of > information for storage? Hmmm. Bit sad, that. Can't we grab some 1541 line > somewhere? :-) That would be 8 more lines. It is not a matter of a missing address line, but a matter of missing half the data lines. The AT/IDE bus is 16 bits wide, you know. And there is only one unused 8-bit I/O port on one of the 6522s inside the 1541. > But, I'm also concerned with using the PC for getting out a "neat" d64 image > out of all the GCR mess..... The PC has so much processing power that this should not be an issue. Marko Message was sent through the cbm-hackers mailing list
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