Assume that we have a 16-bit processor which is interfaced to memory of 32-bits wide(word length).
The processor has to perform two read/write operations when it wants to read/write data respectively.How does it know the word length of memory it is interfaced to?
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C
Chris Hills
In article , Harry writes
Homework?
You will not get any help here unless you show your attempt to solve the problem
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Harry
If you know you can reply or else just shut up!
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Tom Lucas
Well I was going to give the benefit of the doubt but this poster's fate is now sealed I fear...
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Chris Hills
Not only will I not help but I suspect, with that sort of response, you will get no help from anyone. Even though there are many here who could answer your question in their sleep.
Had you bothered to look at this newsgroup (or Usenet in general) you would have known that questions of the type you originally posted get ignored or flamed.
I did neither but constructively suggested how you would get help. By showing how you have attempted a solution.
Don't bite the hand that feeds.
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Chris Hills
I think so too... They never learn.
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\/\/\/\/\ Chris Hills Staffs England /\/\/\/\/
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cs_posting
It doesn't.
It's an issue for the motherboard designer, not the processor vendor, in large part because such a situation would unusual enough that processors are unlikely to have built in support.
What's quite a bit more common, and economically practical, is processors interfaced to half-width memory. For that, you might want to look at the 8086 docs.
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Tim Wescott
If you want to get short shrift on a professional newsgroup, the best way is to pose a problem that's phrased just exactly like a homework problem.
The second best way is to be rude to people who do normally accepted newsgroup things, like telling you they won't do your homework.
Those of us that are professionals don't want to work with or for people who got their degrees fraudulently. Those of us who are students don't want to compete for grades with people who do so fraudulently.
If it's _not_ homework, then say so, and back it up with a comment on your situation. Most of us will be understanding.
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com
Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
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Grant Edwards
You have to tell it. Or not -- it may already know.
Well, that's certainly not going to help.
Grant Edwards grante Yow! I'm definitely not
at in Omaha!
visi.com
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Vladimir Vassilevsky
You have two legs. Car has four wheels. How do you manage to drive the car?
VLV
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Chris Hills
In article , Vladimir Vassilevsky writes
Call a Taxi?
PS if you know say so else shut up! :-)
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Jim Stewart
No, you shut up!! :-)
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Ali
I don't know!
Maybe I've to be bit polite so 2to4 just might work out of blues.
ali
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Chris Hills
I said it first! :-)
(Don't you just love intellectual conversations :-)
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Bob
Bi-quad filter?
Bob
H
Harry
I had to use that statement because why you think that it is a home work or even if it is a home work why don't you people help even though i am not a student.
Anyways I appologize to everyone for behaving so rude........I am so sorry.
It's just a discussion and it.s not a homw work or interview question as you all think.
P
Paul Burke
Look up the instruction set for the processor. You'll find it has instructions something like mov.b, mov.w etc., and maybe mov.l for 8 bit, 16 bit and 32 bit accesses. So the programmer decides what word length to use, while the hardware decides the 16 bit (and maybe the 32 bit) byte order. If it hasn't got long operators, you are free to choose how you access the 32 bits. But probably the compiler (which has a good memory for object sizes) will choose for you.
This was an issue in the old days when we designed with 68000s. That had a 16 bit data bus, with no steering logic to make 8 bit accesses all appear on the lower half of the bus. I knew of at least one STE bus (that shows how long ago it was) 68000 CPU board that could only access odd (or was it even?) addresses through the bus.
Paul Burke
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