wain | WebAssembly implementation from scratch in Safe Rust | Binary Executable Format library
kandi X-RAY | wain Summary
kandi X-RAY | wain Summary
[wain][proj] is a WebAssembly INterpreter written in Rust from scratch with zero dependencies. An implementation of [WebAssembly][wasm-spec]. . Note that this project is in progress. Before v1.0.0 means experimental. Not all of the features are implemented yet. Current status is that all the MVP implementations have been done and many tasks are remaining. Please see [the task board] for current progress. This project started for fun and understanding Wasm deeply.
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QUESTION
I have a python quiz made and it is around 500 lines. i was wondering how i could make it shorter, and simplify the code. This is an example of one of my questions in the quiz
...ANSWER
Answered 2020-Oct-22 at 03:42Below is an example of a custom class that defines a "question" - you could then make many of them, and reuse a lot of code, this way.
QUESTION
My problem:
Lets say that i have class A
with some variable a
And class B
with variables prev
and next
In class A
i want to make method changeIfEqual(B myB)
which checks if A.a == my_B.prev
, if so i change A.a
to my_B.next
. But if A.a != my_B.prev
i want thread to wait()
until continion is true and then execute the thread that have been waining for the longest time.
So I imagine A.changeIfEqual(B myB)
should look like this:
ANSWER
Answered 2019-Nov-12 at 16:45You don’t. Which thread gets notified is up to the scheduler. If you replace the implicit locking (using synchronized) with ReentrantLock, then you can specify that the lock is fair. But that's not a perfect solution, see the API docs:
The constructor for this class accepts an optional fairness parameter. When set true, under contention, locks favor granting access to the longest-waiting thread. Otherwise this lock does not guarantee any particular access order. Programs using fair locks accessed by many threads may display lower overall throughput (i.e., are slower; often much slower) than those using the default setting, but have smaller variances in times to obtain locks and guarantee lack of starvation. Note however, that fairness of locks does not guarantee fairness of thread scheduling. Thus, one of many threads using a fair lock may obtain it multiple times in succession while other active threads are not progressing and not currently holding the lock. Also note that the untimed tryLock method does not honor the fairness setting. It will succeed if the lock is available even if other threads are waiting.
QUESTION
I have In HTML Table but in this table I want remove the duplicates values in each and each
data, See my below my source code
ANSWER
Answered 2019-Jun-27 at 16:20 in this case is a parent element containing HTML, not text. To compare the html contents of multiple
elements, I would suggest using the
.html()
method instead of .text()
.
QUESTION
I have the following table
...ANSWER
Answered 2017-Dec-26 at 11:29First You have to learn Normalization. Database normalization, or simply normalization, is the process of organizing the columns (attributes) and tables (relations) of a relational database to reduce data redundancy and improve data integrity. ... Informally, a relational database relation is often described as "normalized" if it meets third normal form.
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QUESTION
I am very new in Python and I have the following problem.
I have a couple of questions that I have found quite difficult to solve by myself.
Suppose we have two pandas data frames, from which I can extract two columns. Say that in one column, called col1, we have names of subjects, which can be reiterated along the same column, in the following form: 'QUENTIN TARANTINO' or 'Tarantino Quentin'. In the other column, col2, I have entries of the following form: 'TARANTINO, QUENTIN' or 'Tarantino, Quentin', which can be reiterated as well.
Obviously, 'TARANTINO, QUENTIN' or 'Tarantino, Quentin' are the same subject, but are different objects!
I would like to 'merge' these two columns in one column so that the same subject has one and only one label. How can I make it using a list comprehension (or loop) so that it reports both the names (in this form: 'TARANTINO, QUENTIN') and the counts of them?
...ANSWER
Answered 2017-Sep-09 at 13:51The simple way to solve this this is to use set try this : as you need to apply lower to all element in your list first
QUESTION
I have a project which pulls in two local packages. My stack.yaml
has:
ANSWER
Answered 2017-Aug-15 at 18:57From the stack ghci
docs (https://docs.haskellstack.org/en/stable/ghci/):
Similarly to stack build, the default is to load up ghci with all libraries and executables in the project.
So, to load just one package you need to do stack ghci creatur-wains
.
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