I am working on a practice exam question
a
The Problem A stack machine has the input stream of ABCD. Is it possible for an output stream of CDBA? For CDAB? Show the sequence of operations to create these streams or if not possible explain why it is not possible.
I said both output streams were possible.
For CDBA, here are my sequence of operations to create such a stream
push A
push B
push C
pop C/Store
push D
For CDAB
push A
pop A/store
push B
push A
push C
pop C / store
push D
push C
I assumed the stack machine works like a stack does -push pop. Does everyone agree with my two sequences of operations? Is there some property of a "stack machine" that makes these streams impossible to create/produce?
Yup, everything looks good except you need to push C again at then end of CDBA.
If anyone's still curious, I revisited this problem a couple of days ago.
Basically you take in input into the stack machine via push and output via pop(the two principal operations)
So taking a look at our first input stream ABCD, here would be the sequence of operations to produce an output stream of CDBA
Push A
Push B
Push C
Pop C (Output Stream: C)
Push D
Pop D (Output Stream: CD)
Pop B (Output Stream: CDB)
Pop A (Output Stream: CDBA)
And thus an output stream of CDBA is formed
However to produce an output stream of CDAB, the first 6 operations from producing CDBA would apply to get an output stream of CD. However after those 6 operations, you were left with(from top to botto) ['B','A'] which means with just using push and pop, there is no way to output the A before the B.
Therefore the second output stream is not possible to form.
Related
I have one machine with closed protocol and another device "gateway Modbus" from the same manufacter. This gateway convert this protocol to RS-485 Modbus.
When I send a command packet (modbus function 16) to gateway, gateway send (converted) specific packet to the machine and when I inject this packet over simple UART communication, machine can understand and change values too. I create a list with some cloned commands, but I need to know how CRC/checksum/etc is calculed (I think) to create custom packets.
I already used RevEng tool (https://reveng.sourceforge.io/) and CRCcalculator (https://crccalc.com/) trying to find some common crc algorithm with cloned packets, but none worked.
Some cloned packets, where 2 last bytes is CRC/etc. In this packets I changed the temperature value from 0x11 to 0x15 and last 2 bytes changed too (maybe crc/checksum/etc):
9A56F1FE0EB9001100000100641C
9A56F1FE0EB90012000001006720
9A56F1FE0EB90013000001006620
9A56F1FE0EB9001400000100611C
9A56F1FE0EB9001500000100601C
RevEng output:
./reveng -w 16 -l -s 9A56F1FE0EB9001100000100641C 9A56F1FE0EB90012000001006720 9A56F1FE0EB90013000001006620 9A56F1FE0EB9001400000100611C 9A56F1FE0EB9001500000100601C
./reveng: no models found
Someone can help me?
It's not a CRC. The second-to-last byte is the exclusive-or sum of the preceding bytes. I'm not sure what the last byte is, but since it is only taking on two different values in your example, it does not appear to be part of a check value. Or if it is, it's a rather ineffective check value algorithm.
I am a bit confused on this homework and the data structure of a stack. Essentially the question is:
"Give the output printed by java Stack for the input"
Below is the input
it was - the best - of times - - - it was - the - -
the output would be in the answer key
was best times of the was the it (1 left on stack)
I don't understand how you would receive this output I attempted writing my own stack structure but if I iterate through it I just get first in, first out.
Any help will do I would greatly appreciate it.
It isn't stated explicitly, but looking at the input I assume words are pushed onto the stack, while - signs are pop operations. If that is the case then the operations are
push it, stack contains it;
push was, stack contains it was;
pop prints was, stack contains it;
push the, stack contains it the;
push best, stack contains it the best;
pop prints best, stack contains it the;
push of, stack contains it the of;
push times, stack contains it the of times;
pop prints times, stack contains it the of;
pop prints of, stack contains it the;
pop prints the, stack contains it;
push it, stack contains it it;
push was, stack contains it it was;
pop prints was, stack contains it it;
push the, stack contains it it the;
pop prints the, stack contains it it;
pop prints it, stack contains it.
The words in bold are the output of the pop operations.
All,
I am a new user here, and thought I would see if the experts could help me with something I am new to.
I have been given the following statement to try and solve:
The Variable Frequency Drive (VFD) is connected to the PLC by RS485 communication. The speed of the motor (M2) can be adjusted by sending the following command:
STX N DATA ETX , with each separate value having the <> symbols around them.
Data : Length of data is 1 byte, in which the value of S (Slow), M (Medium) or F (Fast) can be sent.
N : Node number of the VFD, with a data length of two byte ASCII.
My question is, how would I type to send this data? It doesn't say whether to use a specific data type to represent, so surely I could just type the data as it is, e.g. STX 1 S ETX?
Othwerside, I'm not sure how to combine the byte representations of the data, representing them in hex, binary or decimal. I'm not sure what is meant by two byte ASCII, is this not UNICODE-16? Also, I'm not sure if I need to send the values of STX or ETX with the data string or not
I hope someone can shed some light on this.
Thanks in advance.
Since the frequency goes from 0-50 Hz, I think we should send data in this range.
So if we want the frequency to be half maximal, we will send 25.
To send this to VFD, we first need to split that number into 2 and 5
The message should read STX 2 5 ETX?
Now we look at the ASCII code table and find 2 and 5.
0x50 = 2
0x53 = 5
We convey everything in a message that reads
STX 0x50 0x53 ETX
The aforementioned S7-300 is recommended for operation. You can also solve this through his TIA portal.
All,
I managed to figure this out with a bit of digging. I simulated it using Siemens S7-300 on TIA portal, and set up communications on a module. I sent the values I wanted using a "move" block, to a value set in the Data Block.
I repeated this for the Node value, making sure the correct data type was chosen, and sent the data through a Send_ptp command block.
Must have got a bit flustered and tired the other night when I was trying it. Hopefully it might help someone in the future.
I am trying to read and write data on Mifare DESFire cards using ISO 7816-4 APDU commands. I followed the steps:
Select application using {00,A4,04,00,07,D2,76,00,00,85,01,00}.
I get the response as 9100.
Then I select file using {00,A4,00,00,02,3F,00} and I get the response as 9100.
Then I try to read the file using command {00,B0,00,00,03} and I get an error with code 6A86.
Now I want to know the exact commands to read and write the data on the cards. I think I still don't get the meaning of P1 and P2 in read/write commands. I already tried many other combinations but I always get errors like 6A86, 6982, etc.
I assume you receive the status word 9000 on success. If you actually receive 9100 when using ISO/IEC 7816-4 basic inter-industry command APDUs something very strange must be going on. 91xx status words should only be returned when you use the ISO/IEC 7816-4 wrapped command set (i.e. when the class byte of the command APDU is set to 0x90).
Let's look at what you currently do:
You start my selecting the MIFARE DESFire ISO AID D2760000850100:
00 A4 0400 07 D2760000850100
This makes sure that the card is in ISO/IEC 7816-4 framing mode and automatically selects the MF (master file; i.e. the PICC level application).
You then, again(!), select the MF (PICC level application):
00 A4 0000 02 3F00
So you are now at the same level as after step 4.
Finally, you try to read 3 bytes starting at offset 0 from the currently selected file using the READ BINARY APDU:
00 B0 0000 03
Since the currently selected file is the MF you can't read binary data from it. The MF does not have a data part. Consequently, the card reports the error code 6A86 ("wrong parameter P1 and/or P2") as it tries to tell you that reading at offset 0 (P1|P2 = 0x0000) from the MF (currently selected file indicated by the upper bit of P1 being zero) makes no sense.
Therefore, before you can read binary data, you need to select an elementary file (EF), probably even located in a different application (dedicated file, DF), that contains an actual data part. This can be done bei either explicitly sending SELECT commands for EF (and, if necessary, the DF) or by implicitly selecting the EF using a short file identifier in P1 of the first read binary command. In the latter case, you would set the upper bit of P1 to one to indicate that the remaining part of P1 encodes a short file ID. In both cases you would need to know the file/application identifiers of the files and applications that you want to read from.
I'm writing a program for a 8051 microcontroller. In the first part of the program I do some calculations and based on the result, I either light the LED or not (using CLR P1.7, where P1.7 is the port the LED is attached to in the microcontroller).
In the next part of the program I want to retrieve the bit, perhaps store it somewhere, and use it in a if-jump instruction like JB. How can I do that?
Also, I've seen the instruction MOV C, P1.7 in a code sample. What's the C here?
The C here is the 8051's carry flag - called that because it can be used to hold the "carry" when doing addition operations on multiple bytes.
It can also be used as a single-bit register - so (as here) where you want to move bits around, you can load it with a port value (such as P1.7) then store it somewhere else, for example:
MOV C, P1.7
MOV <bit-address>, C
Then later you can branch on it using:
JB <bit-address>, <label>
Some of the special function registers are also bit addressable. I believe its all the ones ending in 0 or 8. Don't have a reference in front of me but you can do something like setb r0.1. That way if you need the carry for something you dont have to worry about pushing it and using up space on your stack.