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Bytes and BITS for an idiot


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#1 belcant2

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Posted 20 March 2014 - 10:00 PM

I am learning CompTia A+ and I wish to know more about how data runs through the internal hardware. Mostly I know all the parts and what they do. But I feel it is essential to understand what is most basic, data itself and memory. Basically Bytes and Bits. I have ready a lot about it and understand most things. I do feel stupid asking such things which I am sure are easy to others. 8bit, 16bit, 32bit and 64bit represent how many locations the register can access on the memory 2^n, eg.  2^8=256 locations. Now 8bits equals 1kb and here is where I feel so stupid, thats only 1kb and yet a computer can move gigabytes of data.

Some help on this would be wonderful, it will help me understand so much more, even if it's not in the study material, and ease my frustration.  



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#2 TsVk!

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Posted 20 March 2014 - 10:42 PM

Sweet google, ambrosia for the brain, my endless tome of knowledge....

 

https://www.google.com/search?q=understanding+bits+and+bytes&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:unofficial&client=iceweasel-a&channel=fflb



#3 belcant2

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Posted 20 March 2014 - 11:53 PM

I have searched and search Google believe me it's the first place I always look. But i'll check yours

#4 jonuk76

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Posted 22 March 2014 - 03:13 PM

As far as I am aware, the address bus determines how much memory a CPU can address directly.  A 32 bit address bus can address 4294967296 bytes (2^32) of memory (4096 Mb, 4Gb).  Note how this also is the address limit for 32 bit operating systems which don't support PAE.

 

It's important to note that a processors address bus is often a different number of bits to the internal register.  For example, Motorola 68000 CPU's (designed in 1979, used in for example, Sega Genesis and the Amiga) had a 24 bit address bus, which means it can address 16 Mb.  However, the processor was described as 32 bit internally with a 16 bit external bus, and is normally considered a 16 bit CPU.  The Zilog Z80 (used in many early home computers, and still used in embedded applications) is considered an 8 bit CPU, but it had a 16 bit address bus meaning it could address 64kb of memory.

 

8 bits = 1 byte.  There is also the term octet which specifically means 8 bits (there was some historic ambiguity around the term 'byte').  I don't think you need to get too involved in this for the A+ syllabus, at least I don't recall anything too complicated along these lines.  After all, it's aimed at technicians rather than hardware designers or programmers.


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#5 TsVk!

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Posted 22 March 2014 - 04:58 PM

You're a nice guy jonuk, taking the time to type that out. :thumbsup2:



#6 belcant2

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Posted 22 March 2014 - 07:45 PM

TsVk it's right and thank you for your reply. I was aware of the history, that was part of the course. As far as I can understand from your reply it's the addresses themselves that make up the memory (bytes etc), and the bus determines this. Perhaps your right that it's not necessary for me to be concerned about but I can't help it. I think it helps me understand the computer itself.

Once again thanks a lot for your help.
TsVk it's right and thank you for your reply. I was aware of the history, that was part of the course. As far as I can understand from your reply it's the addresses themselves that make up the memory (bytes etc), and the bus determines this. Perhaps your right that it's not necessary for me to be concerned about but I can't help it. I think it helps me understand the computer itself.

Once again thanks a lot for your help.

#7 TsVk!

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Posted 22 March 2014 - 08:26 PM

To "single shot" answer your question about 32 and 64 bit differences, relating to OS's and programs... It is basically the width of the data unit they process, rather than processing speed, which is defined by clock and channels (threads and cores)...

 

To analogize it... picture a freeway

 

Bits = how wide the freeway is

Clock = how fast the cars are travelling

channels (threads/cores) = how many lanes

 

Hope this helps.

 

( i know this is really simple and misses some steps&details, but helps with the overall theory)

 

I just Googled this search for you... this article seems fairly concise.


Edited by TsVk!, 22 March 2014 - 08:33 PM.


#8 belcant2

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Posted 22 March 2014 - 09:05 PM

Once again thanks so much. This post was cleared and I like the link you found, better than what I found, I shall refer to it many times I am sure. Just one last thing, so if it were 8 bits it would be blocks of 8 0's and 1's side? Your analogy was cool. I understand speed as I studied done physics which is why I don't understand WHY I don't get this. Nonetheless everything is much clearer thanks to you. Maybe I can pester you again in the future lol.

#9 TsVk!

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Posted 22 March 2014 - 09:24 PM

Yes you are right, the bits are the lane width... 8 bit is 8 x 1/0's data group size. (blocks)

 

Google gets smarter the more often you click the right links, and which OS you are using... I get different results on my machine to what you do on yours.



#10 belcant2

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Posted 22 March 2014 - 09:57 PM

Finally :). Now my mind will stop bugging me. Despite all the negative advice I have had from friends and family I will go forward and try get into the industry. Thanks for the step in that direction.

#11 TsVk!

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Posted 22 March 2014 - 10:15 PM

IT is my 3rd career.

 

It is an endless fountain of knowledge, and you will never get bored if you are inquisitive.

 

Just keep asking questions :thumbup2:  before you know it you will be answering them.



#12 jonuk76

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Posted 23 March 2014 - 05:35 AM

You're a nice guy jonuk, taking the time to type that out. :thumbsup2:

 

Thanks :) I try to help where I can.

 

 

TsVk it's right and thank you for your reply. I was aware of the history, that was part of the course. As far as I can understand from your reply it's the addresses themselves that make up the memory (bytes etc), and the bus determines this. Perhaps your right that it's not necessary for me to be concerned about but I can't help it. I think it helps me understand the computer itself.

Once again thanks a lot for your help.
 

 

Yes the addressable memory is determined by the processors address bus.  There may be additional practical limits, for example, Core 2 Duo processors, are 64 bit internally and can address 64Gb as shown in the specs (in other words they have a 36 bit address bus).  However, very few motherboards or chip sets used with this processor would allow you to actually fit 64Gb. 

 

Obviously the course book itself should be your guide, and it definitely helps if you have an instructor who is aware of the current syllabus you can refer to as well. But if you want to do your own reading as well to get a deeper understanding then all the better.  64bit OS's have become a lot more prevalent since I did A+ so I would imagine the differences between 32 and 64 bit OS's should be covered in detail.  If you go on to do Network+ (a step up in complexity IMO) then you will get to know your bits and octets as some knowledge of binary is needed, for example for the sections on TCP/IP and subnets :)


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