I'm taking the Nand2Tetris course on Coursera, the one where we translate VM Bytecode to asm, and I'm a bit confused on how the sub opcode is supposed to work in Project 7.
Assuming I've got code that looks like this
push constant 5
push constant 7
sub
I'd expect the output to be 2, but its -2 instead.
7 was the last thing pushed onto the stack, and I'm assuming the first thing popped off it, but it wasn't the first "argument" to sub? Why is that?
This is just the way two-operand stack operations are defined to work. The way to visualize it is that the operator is moved between the two numbers.
This system also makes it much easier to convert from normal notation, since all you have to do is move the operators, not reorder the numbers.
For example, (1+2)/(4-3) becomes 12+43-/
I want to perform an integer division in Octave (I'm using the latest version 5.1.0). I tired doing idivide(5, 2, "round") which should produce 3. The Documentation says for using idivide with round:
Calculate a ./ b with the fractional part rounded towards the nearest integer.
However, the result I get is 2.5000. What am I doing wrong?
Edit: sorry I used the wrong parameter when initially writing the question. Calling the function with fix works perfectly fine (e.g. idivide(5, 2, "fix") returns 2), I get the problem when using round.
Edit: The below is important for Matlab, in Octave the originally posed question (fix) gave the expected answer.
At least one of the input arguments must be an integer, hence this should work
idivide(5, int32(2), 'fix')
But be sure that this is what you want. idivide will return an integer class object which has various consequences if you aren't expecting it.
For example compare with an alternative solution which returns a double
idivide(5, int32(2), 'fix')*0.9
ans =
int32
2
fix(5/2)*0.9
ans =
1.8000
I am all the time trying to determine how to reverse engineer this checksum. It should be a simple one, it is just a checksum for a firmware version of a device. Here are 5 hex-strings:
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
01854000ff1301050600323436393031343430344d45373435304430373132313239ffffffffffffffffffffffffffffffffffffffffffffffffffffff30303132343639303031333133303031485534355f4543455f4456445f535f4e00443520302e3120ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffeb57
It looks like the last 2 bytes (4 hex-letters) are the checksum. I marked the differences in black.
Is anyone able to find out the algorithm, how the checksum is created? I tried already many things to find it out, but either I did it wrong or it didn´t work.
My guess:
checksum(data) = CRC16-CCITT(data) XOR 0x6155
(which may be equivalent to another standard CRC16, I don't know)
See here for an online demo
Well it can be just about anything.. there are multiple implementation of crc, check for example these, I would apply those crcs on the data and compare their outputs to what you have ..
So say I have a variable, which holds a song number. -> song_no
Depending upon the value of this variable, I wish to call a function.
Say I have many different functions:
Fcn1
....
Fcn2
....
Fcn3
So for example,
If song_no = 1, call Fcn1
If song_no = 2, call Fcn2
and so forth...
How would I do this?
you should have compare function in the instruction set (the post suggests you are looking for assembly solution), the result for that is usually set a True bit or set a value in a register. But you need to check the instruction set for that.
the code should look something like:
load(song_no, $R1)
cmpeq($1,R1) //result is in R3
jmpe Fcn1 //jump if equal
cmpeq ($2,R1)
jmpe Fcn2
....
Hope this helps
I'm not well acquainted with the pic, but these sort of things are usually implemented as a jump table. In short, put pointers to the target routines in an array and call/jump to the entry indexed by your song_no. You just need to calculate the address into the array somehow, so it is very efficient. No compares necessary.
To elaborate on Jens' reply the traditional way of doing on 12/14-bit PICs is the same way you would look up constant data from ROM, except instead of returning an number with RETLW you jump forward to the desired routine with GOTO. The actual jump into the jump table is performed by adding the offset to the program counter.
Something along these lines:
movlw high(table)
movwf PCLATH
movf song_no,w
addlw table
btfsc STATUS,C
incf PCLATH
addwf PCL
table:
goto fcn1
goto fcn2
goto fcn3
.
.
.
Unfortunately there are some subtleties here.
The PIC16 only has an eight-bit accumulator while the address space to jump into is 11-bits. Therefore both a directly writable low-byte (PCL) as well as a latched high-byte PCLATH register is available. The value in the latch is applied as MSB once the jump is taken.
The jump table may cross a page, hence the manual carry into PCLATH. Omit the BTFSC/INCF if you know the table will always stay within a 256-instruction page.
The ADDWF instruction will already have been read and be pointing at table when PCL is to be added to. Therefore a 0 offset jumps to the first table entry.
Unlike the PIC18 each GOTO instruction fits in a single 14-bit instruction word and PCL addresses instructions not bytes, so the offset should not be multiplied by two.
All things considered you're probably better off searching for general PIC16 tutorials. Any of these will clearly explain how data/jump tables work, not to mention begin with the basics of how to handle the chip. Frankly it is a particularly convoluted architecture and I would advice staying with the "free" hi-tech C compiler unless you particularly enjoy logic puzzles or desperately need the performance.
It goes without saying that using hard-coded, hex literal pointers is a disaster:
int *i = 0xDEADBEEF;
// god knows if that location is available
However, what exactly is the danger in using hex literals as variable values?
int i = 0xDEADBEEF;
// what can go wrong?
If these values are indeed "dangerous" due to their use in various debugging scenarios, then this means that even if I do not use these literals, any program that during runtime happens to stumble upon one of these values might crash.
Anyone care to explain the real dangers of using hex literals?
Edit: just to clarify, I am not referring to the general use of constants in source code. I am specifically talking about debug-scenario issues that might come up to the use of hex values, with the specific example of 0xDEADBEEF.
There's no more danger in using a hex literal than any other kind of literal.
If your debugging session ends up executing data as code without you intending it to, you're in a world of pain anyway.
Of course, there's the normal "magic value" vs "well-named constant" code smell/cleanliness issue, but that's not really the sort of danger I think you're talking about.
With few exceptions, nothing is "constant".
We prefer to call them "slow variables" -- their value changes so slowly that we don't mind recompiling to change them.
However, we don't want to have many instances of 0x07 all through an application or a test script, where each instance has a different meaning.
We want to put a label on each constant that makes it totally unambiguous what it means.
if( x == 7 )
What does "7" mean in the above statement? Is it the same thing as
d = y / 7;
Is that the same meaning of "7"?
Test Cases are a slightly different problem. We don't need extensive, careful management of each instance of a numeric literal. Instead, we need documentation.
We can -- to an extent -- explain where "7" comes from by including a tiny bit of a hint in the code.
assertEquals( 7, someFunction(3,4), "Expected 7, see paragraph 7 of use case 7" );
A "constant" should be stated -- and named -- exactly once.
A "result" in a unit test isn't the same thing as a constant, and requires a little care in explaining where it came from.
A hex literal is no different than a decimal literal like 1. Any special significance of a value is due to the context of a particular program.
I believe the concern raised in the IP address formatting question earlier today was not related to the use of hex literals in general, but the specific use of 0xDEADBEEF. At least, that's the way I read it.
There is a concern with using 0xDEADBEEF in particular, though in my opinion it is a small one. The problem is that many debuggers and runtime systems have already co-opted this particular value as a marker value to indicate unallocated heap, bad pointers on the stack, etc.
I don't recall off the top of my head just which debugging and runtime systems use this particular value, but I have seen it used this way several times over the years. If you are debugging in one of these environments, the existence of the 0xDEADBEEF constant in your code will be indistinguishable from the values in unallocated RAM or whatever, so at best you will not have as useful RAM dumps, and at worst you will get warnings from the debugger.
Anyhow, that's what I think the original commenter meant when he told you it was bad for "use in various debugging scenarios."
There's no reason why you shouldn't assign 0xdeadbeef to a variable.
But woe betide the programmer who tries to assign decimal 3735928559, or octal 33653337357, or worst of all: binary 11011110101011011011111011101111.
Big Endian or Little Endian?
One danger is when constants are assigned to an array or structure with different sized members; the endian-ness of the compiler or machine (including JVM vs CLR) will affect the ordering of the bytes.
This issue is true of non-constant values, too, of course.
Here's an, admittedly contrived, example. What is the value of buffer[0] after the last line?
const int TEST[] = { 0x01BADA55, 0xDEADBEEF };
char buffer[BUFSZ];
memcpy( buffer, (void*)TEST, sizeof(TEST));
I don't see any problem with using it as a value. Its just a number after all.
There's no danger in using a hard-coded hex value for a pointer (like your first example) in the right context. In particular, when doing very low-level hardware development, this is the way you access memory-mapped registers. (Though it's best to give them names with a #define, for example.) But at the application level you shouldn't ever need to do an assignment like that.
I use CAFEBABE
I haven't seen it used by any debuggers before.
int *i = 0xDEADBEEF;
// god knows if that location is available
int i = 0xDEADBEEF;
// what can go wrong?
The danger that I see is the same in both cases: you've created a flag value that has no immediate context. There's nothing about i in either case that will let me know 100, 1000 or 10000 lines that there is a potentially critical flag value associated with it. What you've planted is a landmine bug that, if I don't remember to check for it in every possible use, I could be faced with a terrible debugging problem. Every use of i will now have to look like this:
if (i != 0xDEADBEEF) { // Curse the original designer to oblivion
// Actual useful work goes here
}
Repeat the above for all of the 7000 instances where you need to use i in your code.
Now, why is the above worse than this?
if (isIProperlyInitialized()) { // Which could just be a boolean
// Actual useful work goes here
}
At a minimum, I can spot several critical issues:
Spelling: I'm a terrible typist. How easily will you spot 0xDAEDBEEF in a code review? Or 0xDEADBEFF? On the other hand, I know that my compile will barf immediately on isIProperlyInitialised() (insert the obligatory s vs. z debate here).
Exposure of meaning. Rather than trying to hide your flags in the code, you've intentionally created a method that the rest of the code can see.
Opportunities for coupling. It's entirely possible that a pointer or reference is connected to a loosely defined cache. An initialization check could be overloaded to check first if the value is in cache, then to try to bring it back into cache and, if all that fails, return false.
In short, it's just as easy to write the code you really need as it is to create a mysterious magic value. The code-maintainer of the future (who quite likely will be you) will thank you.