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Heroku switched the default Node from 14 to 16 in Dec 2021 for the Ruby buildpack .

Heroku updated the heroku/ruby buildpack Node version from Node 14 to Node 16 (see https://devcenter.heroku.com/changelog-items/2306) which is not compatible with the version of Node Sass locked in at the Webpack version you're likely using.

To fix it do these two things:

1. Specify the 14.x Node version in package.json.

This sounds very like a problem that crops up in Haskell from time to time. The problem is one of garbage collection.

There are two ways that this can go. Firstly, the lazy list can be consumed as it is used, so that the amount of memory consumed is limited. Or, secondly, the lazy list can be evaluated in a way that it remains in memory all of the time, with one end of the list pinned in place because it is still being used - the garbage collector objects to this and spends a lot of time trying to deal with this situation.

Haskell can be as fast as C, but requires the calculation to be strict for this to be possible.

I don't entirely understand the code, but it appears to be recursively creating a longer and longer list, which is then evaluated. Do you have the tools to measure the amount of memory that the garbage collector is having to deal with, and how much time the garbage collector runs for?

You have stumbled into a feature/bug in DispatchSemaphore. If you look at the stack trace and jump to the top of the stack, you'll see assembly with a message:

BUG IN CLIENT OF LIBDISPATCH: Semaphore object deallocated while in use

E.g.,

This is because DispatchSemaphore checks to see whether the semaphore’s associated value is less at deinit than at init, and if so, it fails. In short, if the value is less, libDispatch concludes that the semaphore is still being used.

This may appear to be overly conservative, as this generally happens if the client was sloppy, not because there is necessarily some serious problem. And it would be nice if it issued a meaningful exception message rather forcing us to dig through stack traces. But that is how libDispatch works, and we have to live with it.

All of that having been said, there are three possible solutions:

1. You obviously have a path of execution where you are waiting and not reaching the signal before the object is being deallocated. Change the code so that this cannot happen and your problem goes away.

2. While you should just make sure that wait and signal calls are balanced (fixing the source of the problem), you can use the approach in your question (to address the symptoms of the problem). But that deinit approach solves the problem through the use of non-local reasoning. If you change the initialization, so the value is, for example, five, you or some future programmer have to remember to also go to deinit and insert four more signal calls.

   The other approach is to instantiate the semaphore with a value of zero and then, during initialization, just signal enough times to get the value up to where you want it. Then you won’t have this problem. This keeps the resolution of the problem localized in initialization rather than trying to have to remember to adjust deinit every time you change the non-zero value during initialization.

   See https://lists.apple.com/archives/cocoa-dev/2014/Apr/msg00483.html.

3. Itai enumerated a number of reasons that one should not use semaphores at all. There are lots of other reasons, too:

   • Semaphores are incompatible with new Swift concurrency system (see Swift concurrency: Behind the scenes);
   • Semaphores can also easily introduce deadlocks if not precise in one’s code;
   • Semaphores are generally antithetical to cancellable asynchronous routines; etc.

   Nowadays, semaphores are almost always the wrong solution. If you tell us what problem you are trying to solve with the semaphore, we might be able to recommend other, better, solutions.

You said:

However, if the async function returns before deinit is called and the view controller is deinitialized, then signal() is called twice, which doesn't seem problematic. But is it safe and/or wise to do this?

Technically speaking, over-signaling does not introduce new problems, so you don't really have to worry about that. But this “just in case” over-signaling does have a hint of code smell about it. It tells you that you have cases where you are waiting but never reaching signaling, which suggests a logic error (see point 1 above).

zip will return an iterator. Once unpacked, it cannot be unpacked again, it gets exhausted.

Maybe if you want to make sure that only zip objects get converted to list as you said it would work but it would not be efficient, you can check for it type:

I think "without creating temporary objects" is impossible, especially since "everything is an object" in Python.

You could get O(1) space / number of objects if you implement some sorting algorithm yourself, though if you want O(n log n) time and stability, it's difficult. If you don't care about stability (seems likely, since you say you want to sort by a but then actually sort by a, b and c), heapsort is reasonably easy:

Here are some ways to do this.

1) upper.tri

find the length of the longest substring that appears at least twice (the substrings can overlap)

This problem is also commonly known as Longest repeated substring problem. It can be solved in linear time with a suffix tree.

To solve this problem:

1. Add a special character '$' to the given string S),
2. build a suffix tree from S ;
3. the longest repeated substring of S is indicated by the deepest internal node in the suffix tree, where depth is measured by the number of characters traversed from the root.

Time complexity:

• Suffix tree takes O(nlog(k))) time, where k is the size of the alphabet (if k is considered to be a constant, the asymptotic behaviour is linear)
• tree traversal(for finding longest repeated substring) can be done in O(n) time

With data.table:

The first method sends everything to any() whilst the second only sends to any() when there's an odd number, so any() has fewer elements to go through.