discussion | Dynamic comments for static blogs | REST library

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:

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.

The arithmetic mean of two unsigned integers is defined as:

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?

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

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

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

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

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...

(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.

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: