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Old 07-27-2005, 10:11 PM   #1 (permalink)
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Default bits and bytes

ok, I been reading up a little on this topic on howstuffworks and I have a question. I understand that there are 8 bits in a byte. In the article (which you can see here )it said that "With 8 bits in a byte, you can represent 256 values ranging from 0 to 255."

thats where I get lost....
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Old 07-27-2005, 11:30 PM   #2 (permalink)
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it use binary system
255(D) equals to 11111111(B)...
range : 00000000~11111111 (0~255)
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Old 07-27-2005, 11:40 PM   #3 (permalink)
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yeah but what I don't get is why does it only go up to 255?
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Old 07-27-2005, 11:45 PM   #4 (permalink)
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Those 8 bits are broken down like this,

1 2 4 8 16 32 64 128

So you use a 1 to represent that number, like 0 would be 00000000, 16 would be 00001000 and so forth. If you add them all together, the max number is 255.
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Old 07-28-2005, 12:06 AM   #5 (permalink)
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ah I see, that makes alot more sense now.

sorta....

learning a new number system is kind of confusing. I'll get it eventually. Man do I feel stupid. lol
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Old 07-29-2005, 09:57 PM   #6 (permalink)
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Chog is backwards though, it is:

128 64 32 16 8 4 2 1

Binary is based on the number 2, the digit that is furthest to the right is 2 ^ 0, or 1. To the left of that is 2 ^ 1, to the the left of that is 2 ^ 2 and so on.
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Old 07-29-2005, 10:10 PM   #7 (permalink)
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Its easier just to say that, since we start at 0 and not 1, that a number represented by 8 bits can represent numbers up to 2^8th - 1.

In general, any number with N bits can hold numbers as large as 2^N - 1.
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