laziness | Rails plugin to create tests | Testing library
kandi X-RAY | laziness Summary
kandi X-RAY | laziness Summary
Laziness does one thing: if your application throws an unhandled error, it will automatically create a failing test for you to copy into the appropriate test file - or if you’re using RSpec, it’ll write the appropriate spec for you.
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Top functions reviewed by kandi - BETA
- Constructs a spec .
- Get the description for the given action
- Verify that the route is valid
- Test that the action is not valid
- Returns the path for the given path .
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Community Discussions
Trending Discussions on laziness
QUESTION
For a while F# has supported the ability to auto-quote using []
. Is there anything similar for laziness?
e.g.
...ANSWER
Answered 2021-May-04 at 10:41There is nothing like the ReflectedDefinition
attribute for automatically turning things into delayed Lazy<'T>
computations.
You are right that automatically quoting the argument achieves something like this. You could use the (very limited) LeafExpressionConverter.EvaluateQuotation
to do this for some limited kinds of expressions, but as you note, this would be inefficient. The following is a proof of concept though (but you cannot call custom functions in the branches as this uses LINQ expressions):
QUESTION
Method 1
The usual, very fast, and works great.
...ANSWER
Answered 2021-May-06 at 14:33This is an artifact of how ForJoinPool
handles things when you block its inner threads, and how many new ones it spawns. Though, I could probably find the exact lines where this happens, I am not sure it is worth it. For two reasons:
that logic can change
the code inside
ForkJoinPool
is by far not trivial
It seems that for both of us, ForkJoinPool.commonPool().getParallelism()
will return 11
, so I get the same results as you do. If you log ForkJoinPool.commonPool().getPoolSize()
to find out how many active threads is your code using, you will see that after a certain period, it just stabilizes itself at 64
. So the max tasks that can be processed at the same time is 64
, which is on par with the result that you see (those 8 seconds
).
If I run your code with -Djava.util.concurrent.ForkJoinPool.common.parallelism=50
, it is now executed in 2 seconds
, and the pool size is increased to 256
. That means, there is an internal logic that adjusts these kind of things.
QUESTION
I want to be able to do something like this:
...ANSWER
Answered 2021-Apr-25 at 01:36Remember the assignment expression (:=) was introduced in Python 3.8
You don't call
next()
explicitly, the for loop is what you want, and it does keep the laziness of the generator.To implement with a while loop, I think you should use
try/except StopIteration
, but there is no point, just use the for loop,
QUESTION
I understand that many of the names in Haskell are inspired by category theory terminology, and I'm trying to understand exactly where the analogy begins and ends.
The CategoryHask
I already know that Hask
is not (necessarily) a category due to some technical details about strictness/laziness and seq
, but let's put that aside for now. For clarity,
- The objects of
Hask
are concrete types, that is, types of kind*
. This includes function types likeInt -> [Char]
, but not anything that requires a type parameter likeMaybe :: * -> *
. However, the concrete typeMaybe Int :: *
belongs toHask
. Type constructors / polymorphic functions are more like natural transformations (or other more general maps fromHask
to itself), rather than morphisms. - The morphisms of
Hask
are Haskell functions. For two concrete typesA
andB
, the hom-setHom(A,B)
is the set of functions with signatureA -> B
. - Function composition is given by
f . g
. If we are worried about strictness, we might redefine composition to be strict or be careful about defining equivalence classes of functions.
Functor
s are Endofunctors in Hask
I don't think the technicalities above have anything to do with my confusion below. I think I understand it means to say that every instance of Functor
is an endofunctor in the category Hask
. Namely, if we have
ANSWER
Answered 2021-Apr-21 at 06:31You're right that Konst m
isn't quite a constant functor from a category-theory standpoint. But it's very closely related to one!
QUESTION
Getting started with lazy loading. I have an order with order details.
...ANSWER
Answered 2021-Apr-11 at 06:47The problem is the usage of =>
in _OrderDetails. This is what happens:
If you call Order.OrderDetails.Add(orderDetail)
, this will access the OrderDetails
property which in turn accesses the _OrderDetails
property which will do a new Lazy>
. This will finally create the list and you add a new element to the list.
Fine so far.
But if you then check Order.OrderDetails
again, this will once more access the OrderDetails
property which in turn accesses the _OrderDetails
property which will do another new Lazy>
. So you get a fresh list (without the element you just added).
The solution is simple: replace
QUESTION
From Real World Haskell I read
It operates as follows: when a
seq
expression is evaluated, it forces its first argument to be evaluated, then returns its second argument. It doesn't actually do anything with the first argument:seq
exists solely as a way to force that value to be evaluated.
where I've emphasised the then because to me it implies an order in which the two things happen.
From Hackage I read
The value of
seq a b
is bottom ifa
is bottom, and otherwise equal tob
. In other words, it evaluates the first argumenta
to weak head normal form (WHNF). seq is usually introduced to improve performance by avoiding unneeded laziness.A note on evaluation order: the expression
seq a b
does not guarantee thata
will be evaluated beforeb
. The only guarantee given byseq
is that the botha
andb
will be evaluated beforeseq
returns a value. In particular, this means thatb
may be evaluated beforea
. […]
Furthermore, if I click on the # Source
link from there, the page doesn't exist, so I can't see the code of seq
.
That seems in line with a comment under this answer:
[…]
seq
cannot be defined in normal Haskell
On the other hand (or on the same hand, really), another comment reads:
The 'real'
seq
is defined in GHC.Prim asseq :: a -> b -> b; seq = let x = x in x
. This is only a dummy definition. Basicallyseq
is specially syntax handled particularly by the compiler.
Can anybody shed some light on this topic? Especially in terms of:
- What source is right?
- Is
seq
's implementation really not writable in Haskell?- If so, what does it even mean? That it is a primitive? What does this tell me about what
seq
actually does?
- If so, what does it even mean? That it is a primitive? What does this tell me about what
- In
seq a b
isa
guaranteed to be evaluated beforeb
at least in the case thatb
makes use ofa
, e.g.seq a (a + x)
?
ANSWER
Answered 2021-Apr-04 at 18:21Real World Haskell is mistaken, and all the other things you quoted are correct. If you care deeply about the evaluation order, use pseq
instead.
QUESTION
Context Hi all. I'm having problems and I've looked everywhere and I'm just not getting it. I'm following a youtube tutorial on RecycleView (Coding in flow) however, I'm coming from a tutorial on setting up a MainActivity with fragments so I'm trying to do this inside of a fragment. In the comments, people have had similar problems and he replied by saying that it's basically the same except that inside the fragment you should use getView() however that isn't working.
What I've done so far to remedy the problem I've been looking at SO threads (even ones that are from the same video) and I'm just not getting it, none of their solutions is working for me or the responses they got (if any) were very vague and non-helpful. I've also searched all 300+ comments from that youtube video and looked up other tutorials to see if there's a solution. Not relevant to the problem I'm having but just to get the point across that I've not just types this out of laziness. I've spent the whole day on this problem and I have nothing to show for it
What I need the code to do I need the code to function as per the tutorial found here. I don't believe that the onCreateView is the best place for my code as surely none of the views will be created until that has been run so findViewById wouldn't be possible but what can I do now then? where can I put that code so it will run.
My code
PhotosFragment
...ANSWER
Answered 2021-Mar-23 at 19:21You have a return statement on the first line of onCreateView()
, so none of the other code in there will run. Best to move all that other code to an onViewCreated()
override.
QUESTION
I'm looking to try to run a moderately expensive function on a large list of inputs, using part of the output of that function as one of its inputs. The code runs as expected, unfortunately it consumes a large amount of memory in the process (just under 22GiB on the heap, just over 1GiB maximum residency). Here is a simplified example of what I mean:
...ANSWER
Answered 2021-Mar-10 at 13:46You can't do it like this.
The problem is that your showInts
has to traverse the list twice, first to find the longest number, second to print the numbers with the necessary format. That means the list has to be held in memory between the first and second passes. This isn't a problem with unevaluated thunks; it is simply that the whole list, completely evaluated, is being traversed twice.
The only solution is to generate the same list twice. In this case it is trivial; just have two [0..10^7]
values, one for the maximum length and the second to format them. I suspect in your real application you are reading them from a file or something, in which case you need to read the file twice.
QUESTION
I am trying to implement Huffman compression in Haskell for learning purposes. So far I can successfully create the correct tree for a given String and I can get the correct path to a character from a specific tree. The signature of that function is like this:
...ANSWER
Answered 2021-Feb-28 at 22:32Indeed there's a monadic way for doing that, and it's not particularly fancy. This is exactly what the Maybe
applicative is about, thus you can use traverse
:
QUESTION
I don't understand how laziness / eagerness works in Raku. More precisely how to force eagerness.
I get that Infinite List are lazy. What I don't understand is that some List that have an end are lazy, and why the eager method doesn't work in my case.
I have this example
...ANSWER
Answered 2021-Feb-21 at 23:45This has nothing todo with grep
or eager
, but everything to do with the .gist
method that creates a string representation of an object, and which is called by say
. Since gists are intended for human consumption, they are intentionally not always complete: of Iterable
s, .gist
only shows the first 100 elements:
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