jb | A simple and fast JSON API template engine for Ruby on Rails | Application Framework library
kandi X-RAY | jb Summary
kandi X-RAY | jb Summary
According to the benchmark results, you can expect 2-30x performance improvement in development env, and 3-10x performance improvement in production env.
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Trending Discussions on jb
QUESTION
When I scrape websites for all the emails on each website and try to output it, I can get a given data frame which is a list of sets of multiple elements for each website :
...ANSWER
Answered 2022-Apr-15 at 07:34You can write a function to combine the sets through unions and then perform your filtering when casting it into a list:
QUESTION
I updated java from java 16 to java 17 and now my editor won't work. I use intellij and here is the error message
...ANSWER
Answered 2021-Oct-13 at 21:31Current IntelliJ IDEA version requires Java 11 to run. Remove the overrides (idea.jdk
file/environment variables) to use the default bundled JetBrains Runtime.
QUESTION
IDEAL
MODEL small
STACK 100h
jumps
p186
DATASEG
array dw 312, 340, 311, 300
CODESEG
proc example
pusha
mov al ,4 ;number of elements in array
mov bl, 0 ;index
label1:
mov cx, [array + bx] ;here, every second element ch and cl are swapped
inc bl
cmp bl, al
jb label1
popa
ret
endp example
start:
mov ax, @data
mov ds, ax
call example
exit:
mov ax, 4c00h
int 21h
END start
...ANSWER
Answered 2022-Jan-30 at 19:32The elements of array
are words, 2 bytes each, but you only increment your index bx
by 1 on each iteration of your loop. Addresses on x86 are in units of bytes, so on the next loop iteration, you load from the middle of the word, and you end up with one byte from one element and another byte from the other.
Probably simplest to do everything in terms of bytes, instead of elements. (In 32- or 64-bit code you could use the scale addressing mode, mov cx, [array+ebx*2]
, and then keep counting ebx
in elements, but that's not available in 16-bit real mode.)
Also, the high half of bx
is uninitialized. It's probably best to just use 16-bit values for your indices.
I'd rewrite as (untested):
QUESTION
Given are 2 bitmasks, that should be accessed alternating (0,1,0,1...). I try to get a runtime efficient solution, but find no better way then following example.
...ANSWER
Answered 2022-Jan-28 at 02:44This is quite hard to optimize this loop. The main issue is that each iteration of the loop is dependent of the previous one and even instructions in the loops are dependent. This creates a long nearly sequential chain of instruction to be executed. As a result the processor cannot execute this efficiently. In addition, some instructions in this chain have a quite high latency: tzcnt
has a 3-cycle latency on Intel processors and L1 load/store have a 3 cycle latency.
One solution is work directly with registers instead of an array with indirect accesses so to reduce the length of the chain and especially instruction with the highest latency. This can be done by unrolling the loop twice and splitting the problem in two different ones:
QUESTION
So i have to multiply 'a' by a number of 'b' times and I tried to do it like this. I also took some procedures from other questions I found.
...ANSWER
Answered 2022-Jan-16 at 01:51I keep getting the same error as 'Operand types do not match on lines 18 and 19
That's because TASM knows that the a and b variables are byte-sized given that they were defined using the DB
directive, but your instructions mov bx,a
and mov cx,b
are word-sized operations. So, a mis-match.
Your program is using the a and b 'variables' for user input via the DOS.BufferedInput function 0Ah. Read all about this function here.
What your definitions should look like is:
QUESTION
In my application, I am using JNA to use the native code written in C. Where I get the notification from native application in callback.
In callback, I get a pointer and some other data to process. In JNA callback code, I have to use this pointer again to call some other native library code and have to pass that pointer. After that I have to return from this callback.
If I don't call that intermediate native library method from callback, which passes the pointer, it works fine, but if I add this call my application crashes intermittently (mostly after processing few hundred of callback requests, sometimes it run for thousands of callbacks also sucessfully).
This NotificationHook
class objects for which hook is set up in native code is a static variable, as there will be only one hook for the application. And native library call this one by one.
ANSWER
Answered 2022-Jan-10 at 21:50Based on the code you've provided, the problem is the same as other Callback-related questions here: you're losing the native allocation of TRANX
due to Java's garbage collection.
A JNA Structure
consists of two parts: a pointer (to data), and the data itself. You have not provided the native typedef for TRANX
to confirm your JNA mapping, but an instantiated object will have an internal pointer reference, pointing to a 4-byte allocation of memory (the int unused
).
You only show the callback code where TRANX
is already an argument, meaning you've already instantiated it to pass to the callback.
If you allocated it yourself using new TRANX()
or new TRANX(int unused)
, then JNA has
- allocated 4 bytes of native memory
- stored the pointer to it internally
In JNA, the native memory attached to a Structure
is automatically freed as a part of the garbage collection process. This is a common problem with callbacks, as you generally don't control the timing of the callback return, so the following sequence occurs:
- You create the object in Java (allocating the native 4 bytes which the
TRANX
structure tracks the pointer to internally) - You pass the
TRANX
object to the callback - Immediately after passing the object, Java no longer has need for its own object; it is unreachable and thus eligible for garbage collection
- When GC occurs the native 4 bytes are freed as part of the process
- The
TRANX
object in the callback still has the pointer internally, but it now points to memory that is no longer allocated, resulting in theSIGSEGV
(or Invalid Memory Access error, or strange symptoms if the memory is allocated by another thread, or other undefined behavior).
The solution to the problem is to track the memory associated with TRANX
.
- If you are allocating it yourself, keep a reference to the
TRANX
object to prevent it from being unreachable.- This generally requires accessing the
TRANX
structure at some later point after you are sure the callback will have been processed - In JDK9+ a
ReachabilityFence
can be used for this. - In JDK8 you should manipulate the class in some way (e.g., read a value from it, or call toString on it, etc.).
- This generally requires accessing the
- If you are using a native allocation and creating the pointer from the
peer
value returned from the native API, then read the API to determine when that memory is freed.
QUESTION
The code Should work like this: read a character from the input and look if it is a uppercase Letter repeat the reading until a non uppercase character is read in print out all the uppercase characters starting with the newest read in letter for example: A B C E f
E C B A .... my question would be how do i use the stack to find out if every letter has been printed when is the end reached. and if the first read in char is a non uppercase char it should print out an error message
...ANSWER
Answered 2022-Jan-03 at 23:02"How do I use the stack to find out if every letter has been printed when is the end reached?"
You could always push a special character onto the stack before starting, and during your print loop you can check to see if you encounter this special character before printing. If you do, then instead jump to Ende
. I would of course not use a character in the range of uppercase ASCII, as that would cause issues. Perhaps simply push 0x0 onto the stack to start with and check for that after pop eax
in loop
. Something like this:
QUESTION
I need to sort the array in one proc and than the smallest number should swap its place with the first number . than the second smaller number with the second number and so on... My just wrote 00 00 00 00 in ds:0000
...ANSWER
Answered 2021-Dec-09 at 23:50The min proc exhibits these two errors:
- You are using the opposite of the condition that you would need in order to find the minimum.
- Your code will, in its final iteration, treat the $-terminator as if it were an array element. You must check for the terminator before comparing with a following array element because said element might not exist.
What the code does is InsertionSort in the ascending order.
QUESTION
Background Info -
I have a JSON response from an API call, which I am trying to save in a pandas DataFrame, whilst maintaining the same structure, as when I view in a system I have called the data from.
Functions that calls JSON Response -def api_call():
calls the API (Note: url_list
only contains 1x url at present) and saves the response in the api_response
variable, using json.loads(response.text)
ANSWER
Answered 2021-Dec-05 at 11:11I'm not sure this is the best way to unpack a dictionary, but it works:
(It's used to keep the child "metadata" like the id (duplicate), and holding account full name)
QUESTION
I am trying to test for normality of residuals using 2 different ways.
- Using Jarque-Bera test
- Q-Q Plot
I can see different results, for the JB test the probability value is 19.9553 with a probability of 0.00005. Thus, we can't reject the null hypotheses, and this concludes that there is a non-normal distribution of results.
on the other hand, when I plotted the same dataset using Q-Q graph, I could see a partially linear relation, which might point to a normal distribution. Given the size of observations is 62 and the regression model that was used is the OLS model.
Do you think I did something wrong in my assumption?
...ANSWER
Answered 2021-Nov-21 at 20:14The QQ graph does not show that the data are normally distributed. If you would calculate a single indicator from a QQ plot, then you would measure the (positive vertical ) distances of the points to the red reference line and sum them up. In your case, almost all points deviate from the reference line, voting for a non-normal distribution.
A typical QQ plot of normally distributed data has got a large majority of points on the red reference line and some points at the ends (left and right) may deviate.
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