counter | Realtime counter with node.js , socket.io and smoothie.js | Websocket library
kandi X-RAY | counter Summary
kandi X-RAY | counter Summary
An example of a realtime counter with node.js, socket.io and smoothie.js.
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- Construct a new time series .
- Constructs a new version of the chart
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QUESTION
I'm trying to solve this bonus question from the "How Cairo Works" tutorial. I ran the following function, opened the Cairo tracer and saw that the memory is full with powers of 2. Why is that?
...ANSWER
Answered 2022-Mar-17 at 15:43Here are some leading questions that can help you reach the answer. Answers to the questions after a break:
- Where does the
jmp rel -1
instruction jump to? - What does the target instruction do? What happens after it?
- How did this instruction end up in the program section of the memory?
jmp rel -1
is encoded in the memory at addresses 5-6. When it is executed, we havepc = 5
, thus after the jump we will execute the instruction atpc = 4
, which is0x48307fff7fff8000
.- This bytecode encodes the instruction
[ap] = [ap - 1] + [ap - 1]; ap++
(to check, you can manually decode the flags and offsets, or simply write a cairo program with this instruction and see what it compiles to). After it is executed,pc
is increased by 1, so we again executejmp rel -1
, and so on in an infinite loop. It should be clear why this fills the memory with powers of 2 (the first 2, at address 10, was written by the[fp + 1] = 2; ap++
instruction). - The instruction
[fp] = 5201798304953761792; ap++
has an immediate argument (the right hand side, 5201798304953761792). Instructions with immediate arguments are encoded as two field elements in the memory, the first encoding the general instruction (e.g.[fp] = imm; ap++
), and the second being the immediate value itself. This immediate value is thus written in address 4, and indeed 5201798304953761792 is the same as0x48307fff7fff8000
. Similarly, the2
at address 2 is the immediate argument of the instruction[fp + 1] = 2
, and the-1
at address 6 is the immediate ofjmp rel -1
.
To summarize, this strange behavior is due to the relative jump moving to an address of an immediate value and parsing it as a standalone instruction. Normally this wouldn't occur, since pc
is incremented by 2 after executing an instruction with an immediate value, and by 1 when executing an instruction without one, so it always continues to the next compiled instruction. The unlabeled jump was necessary here to reach this unexpected program counter.
QUESTION
I have a 2D array. It's perfectly okay to iterate the rows in forward order, but when I do it in reverse, it doesn't work. I cannot figure out why.
I'm using MSVC v143 and the C++20 standard.
...ANSWER
Answered 2022-Mar-15 at 20:06This is very likely a code generation bug of MSVC related to pointers to multidimensional arrays: The std::reverse_iterator::operator*()
hidden in the range-based loop is essentially doing a *--p
, where p
is a pointer type to an int[4]
pointing to the end of the array. Decrementing and dereferencing in a single statement causes MSVC to load the address of the local variable p
instead of the address of the previous element pointed to by the decremented p
, essentially resulting in the address of the local variable p
being returned.
You can observe the problem better in the following standalone example (https://godbolt.org/z/x9q5M74Md):
QUESTION
I've been putting together my own disassembler for Sega Mega Drive ROMs, basing my initial work on the MOTOROLA M68000 FAMILY Programmer’s Reference Manual. Having disassembled a considerable chunk of the ROM, I've attempted to reassemble this disassembled output, using VASM as it can accept the Motorola assembly syntax, using its mot
syntax module.
Now, for the vast majority of the reassembly, this has worked well, however there is one wrinkle with operations that have effective addresses defined by the "Program Counter Indirect with Index (8-Bit Displacement) Mode". Given that I'm only now learning Motorola 68000 assembly, I wanted to confirm my understanding and to ask: what is the proper syntax for these operations?
InterpretationFor example, if I have two words:
...ANSWER
Answered 2022-Feb-27 at 12:17In my opinion, both
QUESTION
I have installed Android Studio Canary 2020.3.1.22
and trying to run Flutter
project on Apple Silicon(ARM) Mac
. Unfortunately, it is giving me this error when I try to run default
flutter counter app.
Here is the error I am getting:
...ANSWER
Answered 2022-Jan-03 at 06:03Basically, I installed jdk using brew install java
which was not compatible with my current gradle I guess. So
- I uninstalled java first using:
brew uninstall java
- installed
JDK 8 or JDK 11
from azul. - Installed gradle: gradle-6.9-all.zip
When done, everything worked smoothly.
QUESTION
I can't solve a problem. We have an array. If we take a value, the index of it means port ID, and the value itself means the other port ID it is connected to. Need to find the start index of the longest sequential connection to element which value is -1.
I made a graphic explanation to describe the case for the array [2, 2, 1, 5, 3, -1, 4, 5, 2, 3]. On image the longest connection is purple (3 segments).
I need to make a solution by a function getResult(connections)
with a single argument. I don't know how to do it, so i decided to return another function with several arguments which allows me to make a recursive solution.
ANSWER
Answered 2022-Jan-19 at 22:38The code doesn't work completely properly. Would you please explain my mistakes?
You were quite close. The main problem is that the return
keyword in front of the recursive calls terminates the for
loop and the entire f
function prematurely. This will cause it to visit only the nodes on the first possible branch, not all of them.
The other issue is that branches
might be empty at the end of the function, yet you still access [0][0]
. Instead return the entire array from f
, and access the first tuple on in getResult
.
These two small fixes already make the function work1:
QUESTION
I use std::erase_if
to erase half the elements from containers using a captured counter as follows. C++20 compiled with gcc10
ANSWER
Answered 2021-Dec-28 at 07:50remove_if
takes a Predicate. And the standard library requires that a Predicate type:
Given a glvalue
u
of type (possibly const)T
that designates the same object as*first
,pred(u)
shall be a valid expression that is equal topred(*first)
.
Your predicate changes its internal state. As such, calling it twice with the same element will yield different results. That means it does not fulfill the requirements of Predicate.
And therefore, undefined behavior ensues.
QUESTION
I have the following Dockerfile
:
ANSWER
Answered 2021-Dec-05 at 23:05Does it make sense to iterate through layers like this and keep adding files (to some target, does not matter for now) and deleting the added files in case they are found with a .wh prefix? Or am I totally off and is there a much better way?
There is a much better way, you do not want to reimplement (with worse performances) what Docker already does. The main reason is that Docker uses a mount filesystem called overlay2
by default that allows the creation of images and containers leveraging the concepts of a Union Filesystem: lowerdir
, upperdir
, workdir
and mergeddir
.
What you might not expect is that you can reproduce an image or container building process using the mount
command available in almost any Unix-like machine.
I found a very interesting article that explains how the overlay storage system works and how Docker internally uses it, I highly recommend the reading.
Actually, if you have read the article, the solution is there: you can mount
the image data you have by docker inspect
ing its LowerDir
, UpperDir
, WorkDir
and by setting the merged dir to a custom path. To make the process simpler, you can run a script like:
QUESTION
I'm playing with Rank-N-type and trying to type x x
. But I find it counter intuitive that these two functions can be typed in the same way.
ANSWER
Answered 2021-Nov-08 at 16:22The key is that x :: a -> b
is a function that can provide a value of any type, no matter what argument is given. That means x
can be applied to itself, and the result can be applied to x
again, and so on and so on.
At least, that's what it promises the type checker it can do. The type checker isn't concerned about whether or not any such value exists, only that the types align. Neither f
nor g
can actually be called, because no such value of type a -> b
exists (ignoring bottom and unsafeCoerce
).
QUESTION
Consider the following code, running on an ARM Cortex-A72 processor (optimization guide here). I have included what I expect are resource pressures for each execution port:
Instruction B I0 I1 M L S F0 F1.LBB0_1:
ldr q3, [x1], #16
0.5
0.5
1
ldr q4, [x2], #16
0.5
0.5
1
add x8, x8, #4
0.5
0.5
cmp x8, #508
0.5
0.5
mul v5.4s, v3.4s, v4.4s
2
mul v5.4s, v5.4s, v0.4s
2
smull v6.2d, v5.2s, v1.2s
1
smull2 v5.2d, v5.4s, v2.4s
1
smlal v6.2d, v3.2s, v4.2s
1
smlal2 v5.2d, v3.4s, v4.4s
1
uzp2 v3.4s, v6.4s, v5.4s
1
str q3, [x0], #16
0.5
0.5
1
b.lo .LBB0_1
1
Total port pressure
1
2.5
2.5
0
2
1
8
1
Although uzp2
could run on either the F0 or F1 ports, I chose to attribute it entirely to F1 due to high pressure on F0 and zero pressure on F1 other than this instruction.
There are no dependencies between loop iterations, other than the loop counter and array pointers; and these should be resolved very quickly, compared to the time taken for the rest of the loop body.
Thus, my intuition is that this code should be throughput limited, and considering the worst pressure is on F0, run in 8 cycles per iteration (unless it hits a decoding bottleneck or cache misses). The latter is unlikely given the streaming access pattern, and the fact that arrays comfortably fit in L1 cache. As for the former, considering the constraints listed on section 4.1 of the optimization manual, I project that the loop body is decodable in only 8 cycles.
Yet microbenchmarking indicates that each iteration of the loop body takes 12.5 cycles on average. If no other plausible explanation exists, I may edit the question including further details about how I benchmarked this code, but I'm fairly certain the difference can't be attributed to benchmarking artifacts alone. Also, I have tried to increase the number of iterations to see if performance improved towards an asymptotic limit due to startup/cool-down effects, but it appears to have done so already for the selected value of 128 iterations displayed above.
Manually unrolling the loop to include two calculations per iteration decreased performance to 13 cycles; however, note that this would also duplicate the number of load and store instructions. Interestingly, if the doubled loads and stores are instead replaced by single LD1
/ST1
instructions (two-register format) (e.g. ld1 { v3.4s, v4.4s }, [x1], #32
) then performance improves to 11.75 cycles per iteration. Further unrolling the loop to four calculations per iteration, while using the four-register format of LD1
/ST1
, improves performance to 11.25 cycles per iteration.
In spite of the improvements, the performance is still far away from the 8 cycles per iteration that I expected from looking at resource pressures alone. Even if the CPU made a bad scheduling call and issued uzp2
to F0, revising the resource pressure table would indicate 9 cycles per iteration, still far from actual measurements. So, what's causing this code to run so much slower than expected? What kind of effects am I missing in my analysis?
EDIT: As promised, some more benchmarking details. I run the loop 3 times for warmup, 10 times for say n = 512, and then 10 times for n = 256. I take the minimum cycle count for the n = 512 runs and subtract from the minimum for n = 256. The difference should give me how many cycles it takes to run for n = 256, while canceling out the fixed setup cost (code not shown). In addition, this should ensure all data is in the L1 I and D cache. Measurements are taken by reading the cycle counter (pmccntr_el0
) directly. Any overhead should be canceled out by the measurement strategy above.
ANSWER
Answered 2021-Nov-06 at 13:50First off, you can further reduce the theoretical cycles to 6 by replacing the first mul
with uzp1
and doing the following smull
and smlal
the other way around: mul
, mul
, smull
, smlal
=> smull
, uzp1
, mul
, smlal
This also heavily reduces the register pressure so that we can do an even deeper unrolling (up to 32 per iteration)
And you don't need v2
coefficents, but you can pack them to the higher part of v1
Let's rule out everything by unrolling this deep and writing it in assembly:
QUESTION
In Blazor, how can I undo invalid user input, without changing the state of the component to trigger a re-render?
Here is a simple Blazor counter example (try it online):
...ANSWER
Answered 2021-Nov-18 at 20:12Here's a modified version of your code that does what you want it to:
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