# discussion | Dynamic comments for static blogs | REST library

## kandi X-RAY | discussion Summary

## kandi X-RAY | discussion Summary

Dynamic comments for static blogs. Use it "server-less" with Firebase, or provide your own RESTful API.

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### Top functions reviewed by kandi - BETA

- Generate S3 random position
- Update a model .
- Generates a unique GUID
- Unbind model on the model

## discussion Key Features

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## Community Discussions

Trending Discussions on discussion

QUESTION

I am having trouble resolving a ReDoS vulnerability identified by `npm audit`

. My application has a nested sub-dependency `ansi-html`

that is vulnerable to attack, but unfortunately, it seems that the maintainers have gone AWOL. As you can see in the comments section of that Github issue, to get around this problem, the community has made a fork of the repo called `ansi-html-community`

located here, which addresses this vulnerability.

Thus, I would like to replace all nested references of `ansi-html`

with `ansi-html-community`

.

My normal strategy of using `npm-force-resolutions`

does not seem to be able to override nested sub-dependencies with a different package altogether but rather only the same packages that are a different version number. I have researched this for several hours, but unfortunately, the only way I have found to fix this would appear to be with yarn, which I am now seriously considering using instead of npm. However, this is not ideal as our entire CI/CD pipeline is configured to use npm.

Does anyone know of any other way to accomplish nested sub-dependency package substitution/resolution without having to switch over to using yarn?

Related QuestionsThese are questions of interest that I was able to find, but unfortunately, they tend to only discuss methods to override package version number, not the package itself.

Discusses how to override version number:How do I override nested NPM dependency versions?

Has a comment discussion about`npm shrinkwrap`

(not ideal):
Other related StackOverflow questions:
...ANSWER

Answered 2021-Oct-29 at 21:01I figured it out. As of October 2021, the solution using `npm-force-resolutions`

is actually very similar to how you would specify it using `yarn`

. You just need to provide a link to the tarball where you would normally specify the overriding version number. Your resolutions section of `package.json`

should look like this:

QUESTION

edit: I changed the title from `complement`

to `converse`

after the discussion below.

In the ** operator** module, the binary functions comparing objects take two parameters. But the

`contains`

function has them swapped.I use a list of operators, e.g. `operator.lt`

, `operator.ge`

.

They take 2 arguments, `a`

and `b`

.

I can say `operator.lt(a, b)`

and it will tell me whether `a`

is less than `b`

.

But with `operator.contains`

, I want to know whether `b`

contains `a`

so I have to swap the arguments.

This is a pain because I want a uniform interface, so I can have a user defined list of operations to use (I'm implementing something like Django QL).

I know I could create a helper function which swaps the arguments:

...ANSWER

Answered 2022-Mar-30 at 00:02If either of them posts an answer, you should accept that, but between users @chepner and @khelwood, they gave you most of the answer.

The complement of `operator.contains`

would be something like `operator.does_not_contain`

, so that's not what you're looking for exactly. Although I think a 'reflection' isn't quite what you're after either, since that would essentially be its inverse, if it were defined.

At any rate, as @chepner points out, `contains`

is not backwards. It just not the same as `in`

, `in`

would be `is_contained_by`

as you defined it.

Consider that `a in b`

would not be `a contains b`

, but rather `b contains a`

, so the signature of `operator.contains`

makes sense. It follows the convention of the function's stated infix operation being its name. I.e. `(a < b) == operator.lt(a, b)`

and `b contains a == operator.contains(b, a) == (a in b)`

. (in a world where `contains`

would be an existing infix operator)

Although I wouldn't recommend it, because it may cause confusion with others reading your code and making the wrong assumptions, you could do something like:

QUESTION

When trying to run the command using nextjs npm run dev shows error - failed to load SWC binary see more info here: https://nextjs.org/docs/messages/failed-loading-swc.

I've tried uninstalling node and reinstalling it again with version 16.13 but without success, on the vercel page, but unsuccessful so far. Any tips?

Also, I noticed it's a current issue on NextJS discussion page and it has to do with the new Rust-base compiler which is faster than Babel.

...ANSWER

Answered 2021-Nov-20 at 13:57This worked as suggeted by nextJS docs but it takes away Rust compiler and all its benefits... Here is what I did for those who eventually get stuck...

Step 1. add this line or edit next.json.js

QUESTION

The arithmetic mean of two unsigned integers is defined as:

...ANSWER

Answered 2022-Mar-08 at 10:54The following method avoids overflow and should result in fairly efficient assembly (example) without depending on non-standard features:

QUESTION

I need to calculate the square root of some numbers, for example `√9 = 3`

and `√2 = 1.4142`

. How can I do it in Python?

The inputs will probably be all positive integers, and relatively small (say less than a billion), but just in case they're not, is there anything that might break?

**Related**

- Integer square root in python
- Is there a short-hand for nth root of x in Python?
- Difference between **(1/2), math.sqrt and cmath.sqrt?
- Why is math.sqrt() incorrect for large numbers?
- Python sqrt limit for very large numbers?
- Which is faster in Python: x**.5 or math.sqrt(x)?
- Why does Python give the "wrong" answer for square root? (specific to Python 2)
- calculating n-th roots using Python 3's decimal module
- How can I take the square root of -1 using python? (focused on NumPy)
- Arbitrary precision of square roots

_{Note: This is an attempt at a canonical question after a discussion on Meta about an existing question with the same title.}

ANSWER

Answered 2022-Feb-04 at 19:44`math.sqrt()`

The `math`

module from the standard library has a `sqrt`

function to calculate the square root of a number. It takes any type that can be converted to `float`

(which includes `int`

) as an argument and returns a `float`

.

QUESTION

Discussion about this was started under this answer for quite simple question.

ProblemThis simple code has unexpected overload resolution of constructor for `std::basic_string`

:

ANSWER

Answered 2022-Jan-05 at 12:05Maybe I'm wrong, but it seems that last part:

QUESTION

Originally this is a problem coming up in mathematica.SE, but since multiple programming languages have involved in the discussion, I think it's better to rephrase it a bit and post it here.

In short, michalkvasnicka found that in the following MATLAB sample

...ANSWER

Answered 2021-Dec-30 at 12:23`tic`

/`toc`

should be fine, but it looks like the timing is being skewed by memory pre-allocation.

I can reproduce similar timings to your MATLAB example, however

On first run (

`clear`

workspace)- Loop approach takes 2.08 sec
- Vectorised approach takes 1.04 sec
- Vectorisation saves 50% execution time

On second run (workspace not cleared)

- Loop approach takes 2.55 sec
- Vectorised approach takes 0.065 sec
- Vectorisation "saves" 97.5% execution time

My guess would be that since the loop approach explicitly creates a new matrix via `zeros`

, the memory is reallocated from scratch on every run and you don't see the speed improvement on subsequent runs.

However, when `HH`

remains in memory and the `HH=___`

line outputs a matrix of the same size, I suspect MATLAB is doing some clever memory allocation to speed up the operation.

We can prove this theory with the following test:

QUESTION

Last week, I had a discussion with a colleague in understanding the documentation of C++ features on cppreference.com. We had a look at the documentation of the parameter packs, in particular the meaning of the `(optional)`

marker:

(Another example can be found here.)

I thought it means that this part of the syntax is optional. Meaning I can omit this part in the syntax, but it is always required to be supported by the compiler to comply with the C++ standard. But he stated that it means that it is optional in the standard and that a compiler does not need to support this feature to comply to the standard. Which is it? Both of these explanations make sense to me.

I couldn't find any kind of explanation on the cppreference web site. I also tried to google it but always landed at `std::optional`

...

ANSWER

Answered 2021-Aug-21 at 20:22It means that particular token is optional. For instance both these declarations work:

QUESTION

(Disclaimer: I'm not 100% sure how codatatype works, especially when not referring to terminal algebras).

Consider the "category of types", something like Hask but with whatever adjustment that fits the discussion. Within such a category, it is said that (1) the initial algebras define datatypes, and (2) terminal algebras define codatatypes.

I'm struggling to convince myself of (2).

Consider the functor `T(t) = 1 + a * t`

. I agree that the initial `T`

-algebra is well-defined and indeed defines `[a]`

, the list of `a`

. By definition, the initial `T`

-algebra is a type `X`

together with a function `f :: 1+a*X -> X`

, such that for any other type `Y`

and function `g :: 1+a*Y -> Y`

, there is exactly one function `m :: X -> Y`

such that `m . f = g . T(m)`

(where `.`

denotes the function combination operator as in Haskell). With `f`

interpreted as the list constructor(s), `g`

the initial value and the step function, and `T(m)`

the recursion operation, the equation essentially asserts the unique existance of the function `m`

given any initial value and any step function defined in `g`

, which necessitates an underlying well-behaved `fold`

together with the underlying type, the list of `a`

.

For example, `g :: Unit + (a, Nat) -> Nat`

could be `() -> 0 | (_,n) -> n+1`

, in which case `m`

defines the length function, or `g`

could be `() -> 0 | (_,n) -> 0`

, then `m`

defines a constant zero function. An important fact here is that, for whatever `g`

, `m`

can always be uniquely defined, just as `fold`

does not impose any contraint on its arguments and always produce a unique well-defined result.

This does not seem to hold for terminal algebras.

Consider the same functor `T`

defined above. The definition of the terminal `T`

-algebra is the same as the initial one, except that `m`

is now of type `X -> Y`

and the equation now becomes `m . g = f . T(m)`

. It is said that this should define a potentially infinite list.

I agree that this is sometimes true. For example, when `g :: Unit + (Unit, Int) -> Int`

is defined as `() -> 0 | (_,n) -> n+1`

like before, `m`

then behaves such that `m(0) = ()`

and `m(n+1) = Cons () m(n)`

. For non-negative `n`

, `m(n)`

should be a finite list of units. For any negative `n`

, `m(n)`

should be of infinite length. It can be verified that the equation above holds for such `g`

and `m`

.

With any of the two following modified definition of `g`

, however, I don't see any well-defined `m`

anymore.

First, when `g`

is again `() -> 0 | (_,n) -> n+1`

but is of type `g :: Unit + (Bool, Int) -> Int`

, `m`

must satisfy that `m(g((b,i))) = Cons b m(g(i))`

, which means that the result depends on `b`

. But this is impossible, because `m(g((b,i)))`

is really just `m(i+1)`

which has no mentioning of `b`

whatsoever, so the equation is not well-defined.

Second, when `g`

is again of type `g :: Unit + (Unit, Int) -> Int`

but is defined as the constant zero function `g _ = 0`

, `m`

must satisfy that `m(g(())) = Nil`

and `m(g(((),i))) = Cons () m(g(i))`

, which are contradictory because their left hand sides are the same, both being `m(0)`

, while the right hand sides are never the same.

In summary, there are `T`

-algebras that have no morphism into the supposed terminal `T`

-algebra, which implies that the terminal `T`

-algebra does not exist. The theoretical modeling of the codatatype Stream (or infinite list), if any, cannot be based on the nonexistant terminal algebra of the functor `T(t) = 1 + a * t`

.

Many thanks to any hint of any flaw in the story above.

...ANSWER

Answered 2021-Nov-26 at 19:57(2) terminal algebras define codatatypes.

This is not right, codatatypes are terminal *coalgebras*. For your `T`

functor, a coalgebra is a type `x`

together with `f :: x -> T x`

. A `T`

-coalgebra morphism between `(x1, f1)`

and `(x2, f2)`

is a `g :: x1 -> x2`

such that `fmap g . f1 = f2 . g`

. Using this definition, the terminal `T`

-algebra defines the possibly infinite lists (so-called "colists"), and the terminality is witnessed by the `unfold`

function:

QUESTION

I have a question that is raised from this discussion: C - modify the address of a pointer passed to a function

Let's say I have the following code:

...ANSWER

Answered 2021-Nov-16 at 17:49The postfix increment operator `++`

has higher precedence than the dereference operator `*`

. So this:

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