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Data layout in MC68000 RAM

I take the exam within 8 hours ... I need to do something to make sure I understand correctly before exam MC68000.

question --- Write the following values โ€‹โ€‹in the underlying memory lacquers as the microprocessor will store them as bits or hexadecimal starting at $ 8000

AND

2AC543 ---- for this I need to add two 00s to the right on the right?

5863a04 ------ do i need to add one 0 in front?

5D4 ------ add another 0 right?

AD

BC123 ----- add three 0's right?

FROM

F2

1B4D890378 --- not sure about this part .....

this is how i did it

$ 8000 0A | 00

$ 8002 2A | C5

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1 answer


m68k is big endian, which means the leftmost / first value is the highest. So 0x2AC543 becomes

$8000 0x00 0x2A 0xC5 0x43

An interesting question for 0x5D4: will it be stored as a 16 or 32 bit integer? M68k can do both, therefore, may be correct 0x00 0x00 0x05 0xD4

, and 0x5 0xD4

.



1B.4D89.0378

is obviously too large to store in 32 bits. If you use two long registers for this, you get

$8000 00 00 00 1B   4D 89 03 78

again: the highest order value is needed first.

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