xbytes | Parse bytes to human readable sizes | Parser library
kandi X-RAY | xbytes Summary
kandi X-RAY | xbytes Summary
NodeJS Byte Parser Parse bytes to human readable sizes (4747) → ('4.75 KB') and vice versa.
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Trending Discussions on xbytes
QUESTION
I was attempting to generate a public ECDSA key from a private key, and I haven't managed to find much help on the internet as to how to do this. Pretty much everything is for generating a public key from a public key spec, and I don't know how to get that. So far, this is what I've put together:
...ANSWER
Answered 2018-Mar-13 at 05:22So after a while, I figured out a solution and decided to post it in case anyone else has the same issue as me:
QUESTION
I have an example that I'm attempting to parse. I'm able to get the but cannot get
. I've combined what I've seen in a few other questions on the site here, specifically these:
Unmarshaling XML in Go, golang xml Unmarshal, Golang XML parse
Below is what I've written so far:
...ANSWER
Answered 2019-Nov-14 at 18:49Remove the root level tag:
QUESTION
I have a bitstring which I encoded with the function below when I try to decode with my function it doesn't work. what can I do?
...ANSWER
Answered 2019-May-15 at 14:52(More or less) analogous to how the int
type knows how to convert an integer instance to bytes, it is int
again who knows how to convert bytes back to an integer.
Your xbytes
variable is a bytes
object, so it doesn't know how to convert itself to an integer.
Instead you do it like this:
QUESTION
I use golang generate the P-521 public key. source code look like that:
...ANSWER
Answered 2018-Apr-24 at 16:58secp521r1
uses a 521-bit prime field. So the X or Y coordinates are represented as 521-bit numbers. Yes, 521, not 512.
521 bits is 65 full bytes and one bit left over. In a fixed-size encoding of the public key the most significant byte would always have the 7 most significant bits set to 0, and the 8th bit will be 1 or 0 (so the whole byte is 0x00 or 0x01).
Since the coordinate space isn't quite all of the 521-bit numbers, you have a slightly less than 50% chance of the high bit being set for any particular point.
Go's method seems to be returning the value using a minimum-byte representation. So you should see something like
- 66 bytes: 49% of the time
- 65 bytes: 50% of the time
- 64 bytes (top 9 bits are all 0): 0.2%
- (63 bytes and below are possible, but with vanishingly small percentages)
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