I am learning binary code and I can't get my head around binary negation. I have an example and was hoping someone could walk me through how to answer this question.
Perform binary negation on this binary number. -11011
The answer is 11100101 but I'm not sure how to show the breakdown of how this answer is achieved.
Related
When coding PyTorch in torch.nn.utils I see two functions, clip_grad_norm and clip_grad_norm_.
I want to know the difference so I went to check the documentation but when I searched I only found the clip_grad_norm_ and not clip_grad_norm.
So I'm here to ask if anyone knows the difference.
Pytorch uses the trailing underscore convention for in-place operations. So the difference is that the one with an underscore modifies the tensor in place and the other one leaves the original tensor unmodified and returns a new tensor.
For some backstory, I'm making a program that can do arithmetic on ones complement numbers. To do this I'm converting a binary string into a BigInteger and then performing the math using said BigIntegers, and then converting that back into a binary string. The only problem occurs when the end result goes below -127 or above +127 because I don't know how to correct it due to the nature of ones complement numbers. I was hoping I could somehow instead convert them like unsigned numbers and do like what this answer says to do.
There are also a couple of other questions that I got from reading the linked question. I put them in block quotes. I'm just asking for information on what they mean, and explain it to me.
Firstly
I know that the r-1 complement for r-base number should do end around carry if the highest bit has carry.
Secondly
End-around carry is actually rather simple: it changes the modulus of the addition operation from rn to rn–1.
And lastly
Again, let's keep the carry bit where it is. If you look at the numbers as unsigned integers, we're computing 13 + 11 = 24. However, due to the wrap-around carry, addition is done modulo 15, so we end up with 9, which represents -6 (the correct result).
If someone can explain these quotes to me and provide some web pages for me to read I would greatly appreciate it! :)
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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
01854000ff1131050600323436393031303330324d45363831304330363632333636ffffffffffffffffffffffffffffffffffffffffffffffffffffff30303332343639303036393038303031485534355f4543455f4456445f535f4e00443120302e3120ff7beff9fff36fff7fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffe5d6
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 ..
I am trying to understand the logic behind the arithmetic of one\two complement. Why I can sum\subtract numbers with this method and get the right answer, and why all the issue of "carry" works in this two method as it works?
Someone could help?
I have a homework question regarding operator precedence in Fortran. In order to understand the question I need to know how to use binary numbers in Fortran. Can someone give me an example of how to use binary numbers in fortran? (Specifically with subtraction).
You need to be a bit clearer about what you mean by 'binary numbers in fortran'. In one sense, not terribly useful, all Fortran numbers are binary, as indeed most numbers in most programming languages are binary once they get onto the computer.
Fortran, in the standard at least, does not have the concept of a binary intrinsic data type, it has integers, reals, complex numbers, logicals and characters. Of course, your compiler might implement other types as well, but you don't tell us what that compiler is.
Standard Fortran does have the concept of binary input and output formats -- look for the 'B edit descriptor' in your documentation. This can be used on input and output to read and write binary representations of integers. But the numbers are, to Fortran, integers. So, if you were to read a, b as binary numbers, you would subtract them with the statement a-b.
Fortran does have a set of bit intrinsic procedures, which go by the names iand, ibclr, ieor and so forth but these are really for bit-twiddling.
If you can clarify your questions, I, or some other SOer, might be able to clarify an answer.
Finally, I think it's rather odd that you think you need to know about Fortran 'binary' numbers in order to understand operator precedence. Perhaps you could explain a bit more.