p2 | Push Notifications app server that relays push messages | Notification library
kandi X-RAY | p2 Summary
kandi X-RAY | p2 Summary
An XEP-0357: Push Notifications app server that relays push messages between the user’s server and Googles Firebase Cloud Messaging.
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Top functions reviewed by kandi - BETA
- Register a new command
- Create the response data form
- Creates a new target
- Creates a new MUC target
- Pushes a notification to the given target
- Gets the token
- Create a notification with a content available
- Creates a new alert
- Un - register a command
- Parse a push service node
- Get the singleton TargetStore
- Get device id
- Find the push summary
- Gets the pushSummary value
- Converts value to Jid string
- Validates the service account
- Returns an X509 certificate chain
- Validates the certificate
- Get the push service instance for the given service
- Returns the default trust manager
- Returns the private key for the specified file
- Returns the value for the given Jid parameter
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Trending Discussions on p2
QUESTION
I am solving a problem on LeetCode:
Given an unsorted array of integers
nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs inO(n)time. So fornums=[100,4,200,1,3,2], the output is4.
The Union Find solution to solve this is as below:
...ANSWER
Answered 2022-Mar-14 at 07:33They are right. A properly implemented Union Find with path compression and union by rank has linear run time complexity as a whole, while any individual operation has an amortized constant run time complexity. The exact complexity of m operations of any type is O(m * alpha(n)) where alpha is the inverse Ackerman function. For any possible n in the physical world, the inverse Ackerman function doesn't exceed 4. Thus, we can state that individual operations are constant and algorithm as a whole linear.
The key part for path compression in your code is here:
QUESTION
I am reading this book by Fedor Pikus and he has some very very interesting examples which for me were a surprise.
Particularly this benchmark caught me, where the only difference is that in one of them we use || in if and in another we use |.
ANSWER
Answered 2022-Feb-08 at 19:57Code readability, short-circuiting and it is not guaranteed that Ord will always outperform a || operand.
Computer systems are more complicated than expected, even though they are man-made.
There was a case where a for loop with a much more complicated condition ran faster on an IBM. The CPU didn't cool and thus instructions were executed faster, that was a possible reason. What I am trying to say, focus on other areas to improve code than fighting small-cases which will differ depending on the CPU and the boolean evaluation (compiler optimizations).
QUESTION
struct X { int n; };
const X *p = new const X{3}; // #1
new (const_cast(p)) const X{5}; // #2
const int c = std::launder(p)->n;
ANSWER
Answered 2022-Jan-17 at 18:46[basic.compound]/3 is not relevant. It specifically says that it applies only for the purpose of pointer arithmetic and comparison. There doesn't actually exist an array for the object.
I think when you call std::launder, there are four objects at the relevant address: obj1, obj1.n, obj2 and obj2.n.
obj1 and obj1.n are pointer-interconvertible, as are obj2 and obj2.n. Other combinations aside from identical pairs, are not pointer-interconvertible. There are no array objects and therefore "or the immediately-enclosing array object if Z is an array element." isn't relevant.
When considering reachability from std::launder(p), which points to obj2 thus only obj2 and obj2.n need to be considered as Z in the quote. obj2.n occupies an (improper) subset of bytes of obj2, so it is not relevant. The bytes reachable are those in obj2. Except that I considered obj2.n specifically, this is a rephrasing of your considerations.
By exactly the same reasoning, the bytes reachable from p (pointing to obj1) are all those in obj1.
obj1 and obj2 have the same size and therefore occupy exactly the same bytes. Therefore std::launder(p) would not make any bytes reachable that aren't reachable from p.
QUESTION
With code below (edited basing on code from here) I generates two example tables with gt package:
ANSWER
Answered 2022-Jan-06 at 13:23I can offer to you this solution:
1. We take your data:
QUESTION
This is my code:
...ANSWER
Answered 2021-Dec-21 at 00:17You may find this easier using gridExtra::grid.arrange().
QUESTION
For example,
...ANSWER
Answered 2021-Nov-11 at 05:28var <- structure(character(3), names=letters[1:3])
v2 <- "p2"; names(v2) <- "name p2"
vslice <- function(x, i) x[i]
`vslice<-` <- function(x, i, value){
x[i] <- value
names(x)[i] <- names(value)
x
}
vslice(var, 2)
#> b
#> ""
vslice(var, 2) <- v2
var
#> a name p2 c
#> "" "p2" ""
QUESTION
I made 3 plots with the ggplot2 package. To arrange the plots in a single figure I used the patchwork package. In the arrangement, I put 2 plots at the top, the common legend below these plots and below the common legend the third plot. I created the common legend space with the guide_area() function, but a big unused blank area is created along with it.
How can I keep this unused blank space to a minimum?
...ANSWER
Answered 2021-Oct-12 at 05:59Use heights = ... inside plot_layout.
For example,
QUESTION
Forgive me if this has been answered elsewhere. I've been searching SO and haven't been able to translate the seemingly relevant Q&As to my scenerio.
I'm working on a fun personal project where I have 4 main schemas (barring relationships for now):
- Persona (name, bio)
- Episode (title, plot)
- Clip (url, timestamp)
- Image (url)
Restrictions (Basis of Relationships):
- A Persona can show up in multiple episodes, as well as multiple clips and images from those episodes (but might not be in all clips/images related to an episode).
- An Episode can contain multiple personas, clips, and images.
- An Image/Clip can only be related to a single Episode, but can be related to multiple personas.
- If a Persona is already assigned to episode(s), then any clip/image assigned to the persona can only be from one of those episodes or (if new) must only be capable of having one of the episodes that the persona appeared in associated to the clip/image.
- If an Episode is already assigned persona(s), then any clip/image assigned to the episode must be related to aleast one of those personas or (if new) must only be capable of having one or more of the personas from the episode associated to the clip/image.
I've designed the database structure like so:
This generates the following sql:
...ANSWER
Answered 2021-Oct-05 at 23:19I can't think of any way to add this logic on the DB. Would it be acceptable to manage these constraints in your code? Like this:
Event: a new image would be insterted in DB
QUESTION
I've been willing to dynamically create tab contents in rmarkdown.
I've created an in_tabs that seems to work for everything but ggplot plots.
The way it works is that it creates the Rmd code necessary to display nested lists in tabs.
The following reproducible example shows the issue:
...ANSWER
Answered 2021-Oct-01 at 08:21I'm not 100% sure about all the details so you have to keep in mind that may answer involves some guessing.
When knitting the document knitr runs the ggplot2 code and saves the resulting plot as a png where the filename is the name of the chunk.
As far as I got it from inspecting the md file generated by knitr (by adding keep_md: true to the YAML) the issue with your code is, that "all" plots are saved under the same filename unnamed-chunk-1-1.png, i.e. both of your ggplot chunks look like this in the final md:
QUESTION
I'm working on a pyspark routine to interpolate the missing values in a configuration table.
Imagine a table of configuration values that go from 0 to 50,000. The user specifies a few data points in between (say at 0, 50, 100, 500, 2000, 500000) and we interpolate the remainder. My solution mostly follows this blog post quite closely, except I'm not using any UDFs.
In troubleshooting the performance of this (takes ~3 minutes) I found that one particular window function is taking all of the time, and everything else I'm doing takes mere seconds.
Here is the main area of interest - where I use window functions to fill in the previous and next user-supplied configuration values:
...ANSWER
Answered 2021-Sep-24 at 02:46The solution that doesn't answer the question
In trying various things to speed up my routine, it occurred to me to try re-rewriting my usages of first() to just be usages of last() with a reversed sort order.
So rewriting this:
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