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Is the Jaguar really a 64bit-system?
Q: 'Is the Jaguar really a 64bit-system?'
The biggest problem in this discussion is, that there is NO
definition about what makes a system an 'x-bit' system. Some people
believe that the main processor determines how many bits there are in a
system; if you take the CPU, the Jaguar IS a 64bit-system, because 'TOM'
truely is 64bit and was meant to be the main processor! 'TOM' communicates
via the 64bit system-bus! (=>
see the Jaguar's block-diagram)
Right, but the Motorola 68000 was meant to be ONLY responsible for
boot-strapping, timing, reading the ports, etc. The 68000 wasn't meant to
be used as a real CPU, though many programmers DID use it making the
Jaguar an 'underachiever' in many ways!
isn't of any matter, because ALL RELEVANT communication within the Jaguar
is done in 64bit!
John Mathieson, one of the Jaguar designers:
'Jaguar has a 64-bit memory interface to get a high bandwidth out of cheap DRAM. [...] Where the system needs to be 64 bit then it is 64bit, so the Object Processor, which takes data from DRAM and builds the display is 64bit; and the blitter, which does all the 3D rendering, screen clearing, and pixel shuffling, is 64 bit. Where the system does not need to be 64 bit, it isn't. There is no point in a 64bit address space in a games console! 3D calculations and audio processing do not generally use 64bit-numbers, so there would be no advantage to 64bit processors for this. [...] Jaguar has the data shifting power of a 64bit-system, which is what matters for games, so can reasonably be considered a 64bit-system. But that doesn't mean it has to be 64bits throughout.' (Taken from the Jaguar FAQ)
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