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I stumbled over the following piece of code. The "DerivedFoo" case produces different results on MSVC than on clang or gcc. Namely, clang 13 and gcc 11.2 call the copy constructor of Foo while MSVC v19.29 calls the templated constructor. I am using C++17.

Considering the non-derived case ("Foo") where all compilers agree to call the templated constructor, I think that this is a bug in clang and gcc and that MSVC is correct? Or am I interpreting things wrong and clang/gcc are correct? Can anyone shed some light on what might be going on?

Code (https://godbolt.org/z/bbjasrraj):

I was testing this code (https://godbolt.org/z/fe6hhbeqW)...

Pardon the confusing title.

I have this code, which is accepted by GCC, Clang, and MSVC:

I would like to make a component in SvelteKit which has a randomized parameter. The problem is that the value this parameter takes is different for when the page is rendered server-side versus when that page becomes hydrated.

For example, consider this component:

I am calling slack API(conversations.history) to get channel messages. I want to retrieve sender name in message response. how can i get it?

below is my message response:

I have a Python 3 application running on CentOS Linux 7.7 executing SSH commands against remote hosts. It works properly but today I encountered an odd error executing a command against a "new" remote server (server based on RHEL 6.10):

encountered RSA key, expected OPENSSH key

Executing the same command from the system shell (using the same private key of course) works perfectly fine.

On the remote server I discovered in /var/log/secure that when SSH connection and commands are issued from the source server with Python (using Paramiko) sshd complains about unsupported public key algorithm:

userauth_pubkey: unsupported public key algorithm: rsa-sha2-512

Note that target servers with higher RHEL/CentOS like 7.x don't encounter the issue.

It seems like Paramiko picks/offers the wrong algorithm when negotiating with the remote server when on the contrary SSH shell performs the negotiation properly in the context of this "old" target server. How to get the Python program to work as expected?

Python code

I have this lambda style function call

(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.

In the following program struct S provides two conversion operators: in double and in long long int. Then an object of type S is passed to a function f, overloaded for float and double: