Maximum address Space range that can be utilize in a micro-controller - microcontroller

I have below assignment problem and I do not know my answer is correct or not.
the question is,
"A micro-controller is having only 8 address lines and another additional signal lines through a special register. compute the maximum address space range that can be utilize by the designer ?"
my answer is 2^(16) memory spaces. is my answer is correct? any help is appreciated.

All depends on how many "additional signal lines from a special register" are there.
If that other register gives 8 more lines, without further limitations, and the hardware is correct for such scenario, then yes, your reply is correct, you have 2^16 addresses.
But you didn't state how wide that registers is, and whether that register uses all its bits as address (or paging) lines, or whatever. But generally you are right, the address space width is 2^number_of_lines.

Related

What Are The Reasons For Bit Shifting A Float Before Sending It Via A Network

I work with Unity and C# - when making multiplayer games I've been told that when it comes to values like positions that are floats, I should use a bit shift operator on them before sending them and reverse the operation on receive. I have been told this not only allows for larger numbers values and is capable of maintaining floating point precision which may be lost. However, if I do not have to, I do not wish to run this operation every time I receive a packet unless I have to. Though the bottle necks seem to be the actual parsing of the bytes received. Especially without message framing and attempting to move from string to byte array. (But that's another story!)
My question are:
Are these valid reason to undergo the operation? Are they accurate statements?
If not should I be running bit shift ops on my floats?
If I should, what are the real reasons to do it?
Any additional information would be most appreciated.
One of the resourcesI'm referring to:
Main reasons for going back and forth to/from network byte order is to combat endianness caused problems, mainly to ensure each byte of multi byte values (long, int but also floats) is read and written in the way giving the same results regardless of architecture. This issue can be theoretically ignored if you are sure you are exchanging data between systems using the same endianness, but that's rather bad idea from very beginning as you are simply creating unneded technological debt and keep unjustified exceptions in the code ("It all works BUT on the same endianness only. What can go wrong?").
Depending on your app architecture you can rewrite the packet payload/data once you receive it and then use that version further in the code. Also note that you need to encode the data again prior sending it out.

Dissasemble 68xx code without entry point vector

I am trying to disassemble a code from a old radio containing a 68xx (68hc12 like) microcontroller. The problem is, I dont have the access to the interrupt vector of the micro in the top of the ROM, so I don't know where start to look. I only have the code below the top. There is some suggestion of where or how can I find meaningful routines in the code data?
You can't really disassemble reliably without knowing where the reset vector points. What you can do, however, is try to narrow down the possible reset addresses by eliminating all those other addresses that cannot possibly be a starting point.
So, given that any address in the memory map that contains a valid opcode is a potential reset point, you need to either eliminate it, or keep it for further analysis.
For the 68HC11 case, you could try to guess somewhat the entry point by looking for LDS instructions with legitimate operand value (i.e., pointing at or near the top of available RAM -- if multiple RAM banks, then to any of them).
It may help a bit if you know the device's full memory map, i.e., if external memory is used, its mapping and possible mapped peripherals (e.g., LCD). Do you also know CONFIG register contents?
The LDS instruction is usually either the very first instruction, or close thereafter (so look back a few instructions when you feel you have finally singled out your reset address). The problem here is some data may, by chance, appear as LDS instructions so you could end up with multiple potentially valid entry points. Only one of them is valid, of course.
You can eliminate further by disassembling a few instructions starting from each of these LDS instructions until you either hit an illegal opcode (i.e. obviously not a valid code sequence but an accidental data arrangement that looks like opcodes), or you see a series of instructions that are commonly used in 68HC11 initialization. These involve (usually) initialization of any one or more of the registers BPROT, OPTION, SCI, INIT ($103D in most parts, but for some $3D), etc.
You could write a relatively small script (e.g., in Lua) to do the basic scanning of the memory map and produce a (hopefully small) set of potential reset points to be examined further with a true disassembler for hints like the ones I mentioned.
Now, once you have the reset vector figured out the job becomes somewhat easier but you still need to figure out where any interrupt handlers are located. For this your hint is an RTI instruction and whatever preceding code that normally should acknowledge the specific interrupt it handles.
Hope this helps.

Minimal size of a network buffer

I'm a student and I'm taking right now an Operating Systems course. I've stumbled upon a strange answer for a question while learning for exam and I couldn't find an explanation for it.
Question: Suppose we have an Operating System which runs on low physical memory. Thus the designers decided to make the buffer (that handles all the work that is connected to the network) as small as possible. What can be the smallest size of the buffer?
Answer: Can't be implemented with one byte only, but can be implemented with 2 bytes size.
My thoughts: It has 4 answers, one of them is "3 bytes or more" so I thought that it's the right answer because in order to establish a connection you need at list to be able to send a header of tcp/udp or similar package that contains all the connection info, so I have no idea why it's the right answer (according to the reference). Maybe some degenerate case?
Thanks for help.
The buffer has to be at least as large as the packet size on the network. That will depend upon the type of hardware interface. I know of no network system, even going back to the days of dialup, that used anything close to 2 bytes.
Maybe, in theory, you could have a network system that used 2-byte packets. The same logic would allow you to use 1-byte packets (transmitting fractions of a byte in a packets).
Sometimes I wonder about the questions CS professors come up with. I guess that's why:
Those who can do, do;
Those who can't do, teach;
Those who can't do and can't teach, teach PE.

What does that mean when we say uintptr stores the uninterpreted bits of a pointer value?

I read in golang website that uintptr stores the uninterpreted bits of a pointer value, the kind of anwsers that i found on web were very confusing.Can someone please explaing this to me in simple words .
Short answer: it's used when you need to use an address as if it's a number.
Go is a garbage-collected language. Go always knows exactly, when a thing is a pointer, and when it's just a value. Go needs this knowledge to look for old unused values that it can free. Pointers are also special in Go in that you can't just add a number to a pointer like you would do in C.
But sometimes, when you work with embedded systems or some kind of low-level libraries where you need to add or subtract from an address, you need to tell Go that it shouldn't check this pointer because it's not pointing to anything useful at the moment. You just want to use this address as if it was a number and add/subtract another number to it.
This is where you would use uintptr. This type can hold any pointer value (any address), and when you put an address into it, Go doesn't see it as a pointer, so you can do whatever you want with it.

What is the difference between using mark/space parity and parity none?

What is the purpose having created three type of parity bits that all define a state where the parity bit is precisely not used ?
"If the parity bit is present but not used, it may be referred to as mark parity (when the parity bit is always 1) or space parity (the bit is always 0)" - Wikipedia
There is a very simple and very useful reason to have mark or space parity that appears to be left out here: node address flagging.
Very low-power and/or small embedded systems sometimes utilize an industrial serial bus like RS485 or RS422. Perhaps many very tiny processors may be attached to the same bus.
These tiny devices don't want to waste power or processing time looking at every single character that comes in over the serial port. Most of the time, it's not something they're interested in.
So, you design a bus protocol that uses for example maybe 9 bits... 8 data bits and a mark/space parity bit. Each data packet contains exactly one byte or word (the node address) with the mark parity bit set. Everything else is space parity. Then, these tiny devices can simply wait around for a parity error interrupt. Once it get's the interrupt, it checks that byte. Is that my address? No, go back to sleep.
It's a very power-efficient system... and only 10% wasteful on bandwidth. In many environments, that's a very good trade-off.
So... if you've then got a PC-class system trying to TALK to these tiny devices, you need to be able to set/clear that parity bit. So, you set MARK parity when you transmit the node addresses, and SPACE parity everywhere else.
So there are five possibilities, not three: no parity, mark, space, odd and even. With no parity the extra bit is just omitted in the frame, often selected when the protocol is already checking for errors with a checksum or CRC or data corruption is not deemed likely or critical.
Nobody ever selects mark or space, that's just wasting bandwidth. Modulo some odd standard, like 9-bit data protocols that hardware vendors like to force you to buy their hardware since you have no real shot at reprogramming the UART on the fly without writing a driver.
Setting mark or space parity is useful if you're generating data to send to hardware that requires a parity bit (perhaps because it has a hard coded word length built into the electronics) but doesn't care what its value is.
RS485 requires 9 bits transmission, as described above. RS485 is widely used in industrial applications, whatever the controlled device 'size' (for instance there are many air conditioners or refrigerators offering a RS485 interface, not really 'tiny' things). RS485 allows up to 10Mbs throughput or distances up to 4000 feet. Using the parity bit to distinguish address/data bytes eases hardware implementation, each node of the network can have their own hardware to generate interrupts only if an address byte on the wire matches the node's address.
Very clear and helpful answers and remarks.
For those who find the concept perverse, relax; the term is a problem of semantics rather than information theory or engineering, the difficulty arising from the use of the word "parity".
"Mark" and "space" bits are not parity bits in those applications, and the term arises from the fact that they occupy the bit position in which a parity bit might be expected in other contexts. In reality they have nothing to do with parity, but are used for any relevant purpose where a constant bit value is needed, such as to mark the start of a byte or other signal, or as a delay,or to indicate the status of a signal as being data or address or the like.
Accordingly they sometimes are more logically called "stick (parity) bits", being stuck in "on" or "off" state. Sometimes they really are "don't cares".

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