I am extremely curious to know what the fuck kind of architecture you're running where the width of a memory address is 28-bits.
Sincerely,
FalcoVorbis
Dear 0x0000000:
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Re: Dear 0x0000000:
He is obviously using a notation where the number before the "x" is a bit and the ones after are 9-bit bytes.
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Re: Dear 0x0000000:
But that won't work either. A byte is represented as two characters: 0x00. One character is a nibble (4 bits).
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Re: Dear 0x0000000:
[meta]Man, what a week. Finally found some time to write this...[/meta]
That is because his notation is not hexadecimal. Each digit is 9 bits, so the value goes from 0 to 511. The memory address 1364294724433924810 in binary would be 0001100000111100000110000001100000011000000110000001100002. If we choose UTF-9 as notation:
So, in 0x0000000's "1x63" notation, it would be represented as
That is because his notation is not hexadecimal. Each digit is 9 bits, so the value goes from 0 to 511. The memory address 1364294724433924810 in binary would be 0001100000111100000110000001100000011000000110000001100002. If we choose UTF-9 as notation:
Code: Select all
binary hex 1x63/utf-9
0 0x00 '0'
00110000 0x30 '0'
01111000 0x78 'x'
00110000 0x30 '0'
00110000 0x30 '0'
00110000 0x30 '0'
00110000 0x30 '0'
00110000 0x30 '0'
'0x0000000'I hope that clarifies.
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Re: Dear 0x0000000:
Aaaaaaah... Well fuck, that would work.
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Re: Dear 0x0000000:
Holy shit, I just named myself after an error I got on Windows XP when trying to execute my build. (Just kidding, I meant what bbguimaraes said :P)
Don't ask me about my username.
> viewtopic.php?f=4&t=8520&start=999999#p85581
> viewtopic.php?f=4&t=8520&start=999999#p85581