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06-26-2004, 09:30 AM
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#1
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Member
Registered: Dec 2003
Distribution: Slackware, Alpine Linux, Ubuntu, Debian
Posts: 219
Rep: 
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ASM question
Hy!
I am reading a tutorial about assembly and i don't clearly understand something.
1) 0x80004a8
2) 0x80004ab
3) 0x80004b2
This are memory location addresses. And the writer of the tutorial claims the distance between 1st and 3rd address is 8 bytes, what i can't understand. So please help me out.
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06-26-2004, 10:43 AM
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#2
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Senior Member
Registered: Nov 2002
Location: pikes peak
Distribution: Slackware, LFS
Posts: 2,577
Rep:
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and the answer is ...........Hex
4a8
4a9
4aa
4ab
4ac
4ad
4ae
4af
4b0
4b1
4b2
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06-26-2004, 11:42 AM
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#3
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Member
Registered: Dec 2003
Distribution: Slackware, Alpine Linux, Ubuntu, Debian
Posts: 219
Original Poster
Rep:
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320mb, yes i agree, but why is this sequence 8 bytes long? Why is A(hex) long distance between the memory locations represented by 8 bytes?
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