pentagon | Vault < - > Kubernetes Secrets | Identity Management library
kandi X-RAY | pentagon Summary
kandi X-RAY | pentagon Summary
Apparently, different Vault secrets engines have slightly different APIs for returning data. For instance, here is the response for version 1 of the key/value store:.
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Top functions reviewed by kandi - BETA
- Main entry point .
- getVaultClient returns an API client configured for Vault .
- setVaultTokenViaGCP is used to set a Vault token for a given role
- reconcile reconciles the secrets in the set
- NewReflector creates a new Reflector
- getK8sClient returns a Kubernetes client
- getRoleViaGCP returns the role from GCP metadata
- NewMock returns a mock
- init all engine types
pentagon Key Features
pentagon Examples and Code Snippets
Lines of Code : 85License : Permissive (Apache-2.0)
CopyapiVersion: batch/v1beta1
kind: CronJob
metadata:
name: pentagon
spec:
schedule: "0 15 * * *"
concurrencyPolicy: Replace
jobTemplate:
metadata:
labels:
app: pentagon
spec:
parallelism: 1
completions: 1
{
"request_id": "12a0c057-f475-4bbd-6305-e4c07e66805c",
"lease_id": "",
"renewable": false,
"lease_duration": 2764800,
"data": {
"foo": "world"
},
"wrap_info": null,
"warnings": null,
"auth": null
}
{
"request_id": "78b921ae- vault:
url:
authType: # "token" or "gcp-default"
token: # if authType == "token" is provided
defaultEngineType: # "kv" or "kv-v2" (currently supported)
role: "vault role" # if left empty, queries the GCP metadata service
tls: # optional Community Discussions
Trending Discussions on pentagon
QUESTION
I wrote some code to display a shape (in this case a pentagon) in a window. The pentagon displays just fine on my computer, but when I create a zip file and send it to others to try on their computers the window still opens but the inside of the window is blank rather than displaying a pentagon as it does on my computer. Can anyone tell me why I see a pentagon in the window when I open the exe file on my computer, but others see nothing?
Here is the "main.cpp" code:
...ANSWER
Answered 2022-Mar-06 at 08:09Your vertex and fragment shaders will not compile due to syntax errors.
Vertex shader:
layout (location = 0 in vec3 Apos;
QUESTION
Of the various ways to initialize std::array class members in the constructor initializer list, I found the variadic parameter pack to work best.
However, how can I initialize a second std::array member which depends on the first?
The example here is a struct polygon which takes a variadic parameter pack of Vector2 vertices. From those vertices we also construct the lines. These lines store the vertices again (references or pointers seem unnecessary given the typical 16byte size of a Vector2) in pairs of line.start and line.end and also pre-compute the Length and normal vector of the line for performance.
Note: I don't want to use two std::vectors - that's what the working code has already.
Is there a way to move the logic from the constructor body to the constructor initializer list for this case?
...ANSWER
Answered 2022-Feb-19 at 19:28You can put patterns in front of a ..., not just the pack. However, this requires a suitable pack. In this case, a useful pack would be the indices, so you can make a helper for that:
QUESTION
I want to calculate index of the h3 cells between (-39.08, -57.01) and (-38.05, -58.16) but it throws LineUndefinedException.
In the documentation it says
This function may fail to find the line between two indexes, for example if they are very far apart. It may also fail when finding distances for indexes on opposite sides of a pentagon.
I think that it throws the exception because of the latter case. Is there workaround to calculate those cell indexes?
Code:
...ANSWER
Answered 2022-Feb-08 at 17:41You have hit one of the edge cases H3 does not handle (yet), points on opposite sides of a pentagon:
The image above shows the edges of the icosahedron the H3 grid is based on. Pentagons are located at the vertexes of the icosahedron, and your points cross these edges in a way that makes it harder to compute the grid path.
The correct workaround here, which we eventually intend to implement in the library, is to find the intersection points between the great arc linking the two points and the icosahedron edges. To do this, you'd need the icosahedron geometry, which is available in the H3 lib but slightly hard to extract. There's a version here which has a bunch of unnecessary intermediate points in each edge (to facilitate rendering great arcs correctly on web maps), but which should work. The basic algorithm is:
- Find all intersection points along the great arc between point A and point B.
- For each point, find the H3 index.
- Find the
h3Linebetween each segment, e.g.h3Line(cellA, cellIntersection1),h3Line(cellIntersection1, cellIntersection2), and so on. - Combine the lines, dropping duplicate indexes.
There is likely an easier option here, which is to do the same thing but instead of calculating indexes just sample along the great arc. In this particular case, it looks like you probably could just take the center of the great arc and calculate h3Line twice, but in some cases you'd need more samples.
QUESTION
I am having some trouble understanding all the info in the H3 indexing docs. For those who are new to H3, it is a "Hexagonal hierarchical geospatial indexing system" recently released from Uber. Here is the key bit of info for this question:
H3 Cell Indexβ
An H3 Cell index (mode 1) represents a cell (hexagon or pentagon) in the H3 grid system at a particular resolution. The components of the H3 Cell index are packed into a 64-bit integer in order, highest bit first, as follows:
- 1 bit reserved and set to 0,
- 4 bits to indicate the H3 Cell index mode,
- 3 bits reserved and set to 0,
- 4 bits to indicate the cell resolution 0-15,
- 7 bits to indicate the base cell 0-121,
- 3 bits to indicate each subsequent digit 0-6 from resolution 1 up to the resolution of the cell (45 bits total are reserved for resolutions 1-15)
The three bits for each unused digit are set to 7.
I'm not sure what "indicate the base cell 0-121" means, or the next line "indicate each subsequent digit 0-6" ...
First part of the question (that may help me figure it out on my own) is what those two last bullets mean. But the main question is, is it possible to map an incrementing integer (starting from 0) to every cell within a single resolution level, just based on bit manipulation (i.e. not by using a hash table or map of some sort)? I can't quite tell, and not knowing these last two lines I'm not seeing the approach if it's possible yet.
...ANSWER
Answered 2022-Jan-13 at 23:32This breakdown of the bit layout might help (in terms of understanding the index structure).
- There are 122 base cells. Every cell in the grid is either one of these base cells or a descendant of one of these base cells, so every index contains the base cell it descends from.
- The rest of the index is hierarchical, walking down from the base cell. Essentially an address like "Base cell 27 child 2 child 0 child 3...".
It is possible to do what you're describing, purely with bit manipulation, but it requires a fair amount of knowledge of the H3 indexing system. I've done this before, but the code is currently proprietary and I can't share it here :(. As with a lot of H3 code, the pentagons are the hard part.
The solution looks something like this:
- Determine the cell count at the desired res
rfor pentagon and non-pentagon base cells. Hex cells have7^rchildren, pentagons have1 + 5 * (7^r - 1) / 6children (you can usemaxH3ChildrenSizefor this, but you'll need to know it anyway). - You'll need to know the pentagon base cell numbers. These are
4, 14, 24, 38, 49, 58, 63, 72, 83, 97, 107, 117
For a given number n:
- Start with an "empty" index,
8ffffffffffffff. Set the resolution bits tor. - Determine the base cell
nis in and the base cell offset (numPrecedingHexBaseCells * hexBaseSize + numPrecedingPentBaseCells * pentBaseSize). - Set the base cell bits in the index.
- Subtract the offset from
n. This is your child offset (i.e. the offset of the cell within the list of base cell children). - Use the child offset to figure out the index numbers for res 1, res 2, ...res
r. I can't easily walk through the logic for this here, but you should be able to derive it from the formulas above for calculating the child counts (e.g. at res 1, a hex base cell has 7 children. At res 2, it has 49 - first figure out which res 1 groupnis in withn % 7, then figure out the res 2 number, etc).
This is all pretty complex; we're planning to add it to the library, but it will be a few months at least. If you just need a streaming list of all cells at a res, and the performance isn't a strong requirement, you could use a DFS approach instead - for each base cell, find all children, then find all children of one child at the next res, then one child at the next res, etc until the desired resolution. This would only require keeping 7 * r cells in memory at a time.
QUESTION
I just learned about H3 and would like to basically get the "standard reference" point, referencing each hexagon (and the 12 pentagons) at each resolution, and then determine what the lat/lng coordinates are for each hexagon. How does this work?
The H3 grid is constructed on the icosahedron by recursively creating increasingly higher precision hexagon grids until the desired resolution is achieved. Note that it is impossible to tile the sphere/icosahedron completely with hexagons; each resolution of an icosahedral hexagon grid must contain exactly 12 pentagons at every resolution, with one pentagon centered on each of the icosahedron vertices.
The first H3 resolution (resolution 0) consists of 122 cells (110 hexagons and 12 icosahedron vertex-centered pentagons), referred to as the base cells. These were chosen to capture as much of the symmetry of the spherical icosahedron as possible. These base cells are assigned numbers from 0 to 121 based on the latitude of their center points; base cell 0 has the northern most center point, while base cell 121 has the southern most center point.
Each subsequent resolution beyond resolution 0 is created using an aperture 7 resolution spacing (aperture refers to the number of cells in the next finer resolution grid for each cell); as resolution increases the unit length is scaled by sqrt(7) and each hexagon has 1/7th the area of a hexagon at the next coarser resolution (as measured on the icosahedron). H3 provides 15 finer grid resolutions in addition to the resolution 0 base cells. The finest resolution, resolution 15, has cells with an area of less than 1 m2.
I would like to go down to resolution 10, which is about 15m^2 area. I would like to store a custom generated key with each hexagon at this resolution, associating the key with the index or hash of the hexagon, which I assume is somehow associated to the origin of the system, and somehow the origin is hardcoded to a specific place on the globe.
So it appears that h3.getRes0Indexes will give us the 122 origin cells at base 0. And the hierarchy docs show in the JavaScript library how to traverse the resolutions at least.
ANSWER
Answered 2022-Jan-13 at 17:54where is the origin?
There is no single origin. The grid is based on a projection of a regular icosahedron onto the surface of the globe (using a spherical model).
Where is the part where it maps the stuff to the actual Earth globe? Is that hardcoded somehow in the codebase somewhere?
The orientation of the icosahedron is fixed relative to the Earth - you can see the center coordinates of each face here (these are, as you suggested, hardcoded in the codebase).
How do I know that when I use the library today, the 122 base hexagons I get from the API will have the same hash as the last time I used it?
All H3 indexes are stable - this is a function of the algorithm we use to create the index, and of the API guarantee we offer that we won't make any changes to the library that change cell indexes or geographic locations.
You can see more info on the methodology of the indexing function here.
QUESTION
I have recently started learning Java and currently studying about polymorphism. Consider the following code :
...ANSWER
Answered 2022-Jan-10 at 13:12In Java, a method signature is part of the method declaration. It's the combination of the method name and the parameter list.
When we talk about the same method signature in the superClass and the subClass + a covariant return type , so we are talking about the Overriding .
(Same signature + Covariant return type) == OverridingQUESTION
ANSWER
Answered 2021-Dec-16 at 03:07A possible approach would involve the calculation of some blob descriptors and filter blobs according to those properties. For example, you can compute the blob's aspect ratio, the (approximated) number of vertices and area. The steps are very straightforward:
- Load the image and convert it to grayscale.
- (Invert) Threshold the image. Letβs make sure the blobs are colored in white.
- Get the binary imageβs contours.
- Compute two features: aspect ratio and number of vertices
- Filter the blobs based on those features
Letβs see the code:
QUESTION
I need to create a function in R that takes as input an integer, S β₯ 1 and returns as output the pentagonal number which is closest to S.The output of my function should be the pentagonal number ππ which satisfies |ππβπ |β€|ππβπ | for all positive integers m.
However if I could get two different pentagonal numbers which happens when the integer, s is literally in the middle of them. Then it doesn't matter which one it takes (greater or lesser value) which is like when S is 17 and the pentagonal number closest to 17 is 12 and 22 so it can take either one.
Here is the following code that I have created which is used to find the pentagonal number ππ for a given positive integer, n:
...ANSWER
Answered 2021-Dec-04 at 22:49You don't need loops, just solve following problem:
For input S find minimum n such that: 3n^2-n-2S >= 0
By doing that you get your two candidates:
QUESTION
Good day stackoverflow community. Have a non trivial question. For displaying web content we are using ActiveX. We decided to switch to QWebEngine, but were faced with the problems:
ActiveX allows us:
- save HTML anchors in QAxObject
- listen to anchors click events
- extract and change anchors properties ("id", "href")
Example of code:
...ANSWER
Answered 2021-Sep-14 at 12:45The answer is:
- Use
QWebEnginePage::runJavaScript()if you don't need to listen to the click events. - For listening to click events - use
QWebChanell
QUESTION
I have code that produces the following df as output:
...ANSWER
Answered 2021-Sep-06 at 10:49I dont think your for iterates all the rows.
do this:
for i in range(len(df)):
Also you can remove i = i + 1
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