exu | Universal Executables are an architecture

 by   lachlansneff Rust Version: Current License: MIT

kandi X-RAY | exu Summary

kandi X-RAY | exu Summary

exu is a Rust library. exu has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Universal Executables are an architecture independent, strongly versioned, and statically sandboxed executable format that consists of a WebAssembly module combined with a compressed, read-only filesystem and transactional database. The format of .exu files is described in format.md.
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            kandi-support Support

              exu has a low active ecosystem.
              It has 14 star(s) with 0 fork(s). There are 4 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              exu has no issues reported. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of exu is current.

            kandi-Quality Quality

              exu has no bugs reported.

            kandi-Security Security

              exu has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

            kandi-License License

              exu is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              exu releases are not available. You will need to build from source code and install.

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            exu Key Features

            No Key Features are available at this moment for exu.

            exu Examples and Code Snippets

            No Code Snippets are available at this moment for exu.

            Community Discussions

            QUESTION

            Location and time of impact for a moving point and a moving line segment (Continuous Collision Detection)
            Asked 2020-Aug-21 at 06:04

            I am working on a 2D collider system that breaks up shapes into one possible primitive: impenetrable segments that are defined by two points. To provide collision detection for this system, I am using a static collision detection approach that calculates the distance between the edge of one segment and the currently handled segment (point/line distance) once every frame. If the distance is too small, a collision is triggered during that frame. This works fine but has the known problem of tunneling if one or more bodies exhibit high speeds. So I am tinkering with alternatives.

            Now I want to introduce continuous collision detection (CCD) that operates on dynamic points / dynamic segments. My problem is: I don't exactly know how. I do know how to do continuous collision between two moving points, a moving point and a static segment but not how to do CCD between a moving point (defined by point P) and a moving segment (defined by points U and V, both can move completely freely).

            illustration of problem

            I have seen similar questions beeing asked on SO and other platforms, but not with these exact requirements:

            • both point and segment are moving
            • segment can be rotating and stretching (because U and V are moving freely)
            • collision time and collision point need to be found accurately between two frames (CCD, no static collision test)
            • I prefer a mathematically perfect solutiuon, if possible (no iterative approximation algorithms, swept volumes)
            • note: the swept line shape will not always be a convex polygon, because of the freedom of the U,V points (see image)
            • note: testing for a collision with the swept volume test is inaccurate because a collision point with the polygon does not mean a collision point in the actual movement (see image, the point will have left the polygon once the actual segment has crossed the trajectory of the point)

            So far I came up with the following approach, given:

            • sP (P at start of frame),
            • eP (P at end of frame),
            • sU (U at start of frame),
            • eU (U at end of frame),
            • sV (V at start of frame),
            • eV (V at end of frame)

            Question: Will they collide? If yes, when and where?

            To answer the question of "if", I found this paper to be useful: https://www.cs.ubc.ca/~rbridson/docs/brochu-siggraph2012-ccd.pdf (section 3.1) but I could not derive the answers to "when" and "where". I also found an alternative explanation of the problem here: http://15462.courses.cs.cmu.edu/fall2018/article/13 (3rd Question)

            Solution:

            Model temporal trajectory of each point during a frame as linear movement (line trajectory for 0 <= t <= 1)

            • P(t) = sP * (1 - t) + eP * t
            • U(t) = sU * (1 - t) + eU * t
            • V(t) = sV * (1 - t) + eV * t

            (0 <= a <= 1 represents a location on the segment defined by U and V):

            • UV(a, t) = U(t) * (1 - a) + V(t) * a

            Model collision by equating point and segment equations:

            • P(t) = UV(a, t)
            • P(t) = U(t) * (1 - a) + V(t) * a

            Derive a function for the vector from point P to a point on the segment (see picture of F):

            • F(a, t) = P(t) - (1 - a) * U(t) - a * V(t)

            To now find a collision, one needs to find a and t, so that F(a, t) = (0, 0) and a,t in [0, 1]. This can be modeled as a root finding problem with 2 variables.

            Insert the temporal trajectory equations into F(a, t):

            • F(a, t) = (sP * (1 - t) + eP * t) - (1 - a) * (sU * (1 - t) + eU * t) - a * (sV * (1 - t) + eV * t)

            Separate the temporal trajectory equations by dimension (x and y):

            • Fx(a, t) = (sP.x * (1 - t) + eP.x * t) - (1 - a) * (sU.x * (1 - t) + eU.x * t) - a * (sV.x * (1 - t) + eV.x * t)

            • Fy(a, t) = (sP.y * (1 - t) + eP.y * t) - (1 - a) * (sU.y * (1 - t) + eU.y * t) - a * (sV.y * (1 - t) + eV.y * t)

            Now we have two equations and two variables that we want to solve for (Fx, Fy and a, t respectively), so we should be able to use a solver to get a and t to only then check if they lie within [0, 1].. right?

            When I plug this into Python sympy to solve:

            ...

            ANSWER

            Answered 2020-Aug-21 at 06:04
            TLDR

            I did read "...like 5 minutes to evaluate..."

            No way too long, this is a real-time solution for many lines and points.

            Sorry this is not a complete answer (I did not rationalize and simplify the equation) that will find the point of intercept, that I leave to you.

            Also I can see several approaches to the solution as it revolves around a triangle (see image) that when flat is the solution. The approach bellow finds the point in time when the long side of the triangle is equal to the sum of the shorter two.

            Solving for u (time)

            This can be done as a simple quadratic with the coefficients derived from the 3 starting points, the vector over unit time of each point. Solving for u

            The image below give more details.

            • The point P is the start pos of point
            • The points L1, L2 are the start points of line ends.
            • The vector V1 is for the point, over unit time (along green line).
            • The vectors V2,V3 are for the line ends over unit time.
            • u is the unit time
            • A is the point (blue), and B and C are the line end points (red)

            There is (may) a point in time u where A is on the line B,C. At this point in time the length of the lines AB (as a) and AC (as c) sum to equal the length of line BC (as b) (orange line).

            That means that when b - (a + c) == 0 the point is on the line. In the image the points are squared as this simplifies it a little. b2 - (a2 + c2) == 0

            At the bottom of image is the equation (quadratic) in terms of u, P, L1, L2, V1, V2, V3.

            That equation needs to be rearranged such that you get (???)u2 + (???)u + (???) = 0

            Sorry doing that manually is very tedious and very prone to mistakes. I don`t have the tools at hand to do that nor do I use python so the math lib you are using is unknown to me. However it should be able to help you find how to calculate the coefficients for (???)u2 + (???)u + (???) = 0

            Update

            Ignore most of the above as I made a mistake. b - (a + c) == 0 is not the same as b2 - (a2 + c2) == 0. The first one is the one needed and that is a problem when dealing with radicals (Note that there could still be a solution using a + bi == sqrt(a^2 + b^2) where i is the imaginary number).

            Another solution

            So I explored the other options.

            The simplest has a slight flaw. It will return the time of intercept. However that must be validated as it will also return the time for intercepts when it intercepts the line, rather than the line segment BC

            Thus when a result is found you then test it by dividing the dot product of the found point and line segment with the square of the line segments length. See function isPointOnLine in test snippet.

            To solve I use the fact that the cross product of the line BC and the vector from B to A will be 0 when the point is on the line.

            Some renaming

            Using the image above I renamed the variables so that it is easier for me to do all the fiddly bits.

            Source https://stackoverflow.com/questions/63486253

            QUESTION

            Bash source doesn't bail out on errexit, any possible reason?
            Asked 2020-Jun-16 at 09:40

            I couldn't explain what happens in my scripts, could anyone shed some light please?

            I am doing some pretty standard stuff, set errexit, sourcing one script from another, catching errors and eventually bailing out if any.

            s1.sh

            ...

            ANSWER

            Answered 2020-Jun-16 at 09:40

            The UNBOUND VARIABLE error comes because you are using set -u are referencing with $UNBOUND_VARIABLE a variable named _UNBOUND_VARIABLE_ which has not been assigned to, in the statement

            Source https://stackoverflow.com/questions/62403500

            QUESTION

            Merging two joined tables and output as JSON?
            Asked 2019-May-17 at 12:06

            I want to SELECT entities and an uploads JSON array column containing all of their uploads, along with the cover and profile fields for each upload.

            So far, I can get the array of uploads, but I am unable to add the two other fields from the entitiesXuploads table.

            A basic representation of my three tables is as such:

            Entities

            • id (int8)
            • name (varchar)

            Uploads

            • id (uuid)
            • title (varchar)
            • versions (jsonb)

            entitiesXuploads

            • entityId (int8)
            • uploadId (uuid)
            • cover (bool)
            • profile (bool)

            My query:

            ...

            ANSWER

            Answered 2019-May-17 at 12:06

            QUESTION

            Changing the way my program accepts files will create this error from GSON: Expected BEGIN_ARRAY but was BEGIN_OBJECT at line 1 column 2 path $?
            Asked 2018-Aug-30 at 15:48

            There's been a lot of questions asked about this topic but this question is different because the way I define a certain variable is what determines whether or not this error is thrown.

            Basically, I have the below JSON file:

            ...

            ANSWER

            Answered 2018-Aug-30 at 15:48

            The reason for this problem is that you are referring to two different JSON's

            the one you are passing in string is capsJsonArray.json

            whereas the one you are passing in maven is cfgJsonFile.json

            I think the problem lies in json being refereed as it may be incorrect

            Source https://stackoverflow.com/questions/52100577

            QUESTION

            Update a value of the first column of a file and keep appending n times
            Asked 2017-Mar-29 at 17:52

            I am looking for a better way to add 2 on contents on the first column in a file, and keep appending n times. This can be an one-liner program.

            Input file (sparc_test.export, it is more than 3000 lines in an actual file):

            ...

            ANSWER

            Answered 2017-Mar-29 at 17:52

            The only missing piece is that you don't copy the input file first. (Well, and why C-style for loop?)

            You can also avoid extra file operations, though -- read into an array and work with that.

            Source https://stackoverflow.com/questions/43076022

            QUESTION

            Perl - update contents of a file by adding 1 on the first column
            Asked 2017-Mar-08 at 20:18

            I am looking for a better way to add 1 on contents on the first column in a file. I read a file and update contents on the first column by adding 1.

            my input file: (sparc_export.test)

            ...

            ANSWER

            Answered 2017-Mar-08 at 20:18

            This can be done very easily with just a short command rather than a full program:

            Source https://stackoverflow.com/questions/42679987

            QUESTION

            Perl - Searching values in a log file and store/print them as a string.
            Asked 2017-Feb-25 at 06:10

            I would like to search values after a specific word (Current Value = ) in a log file, and makes a string with values.

            vcs_output.log: a log file

            ...

            ANSWER

            Answered 2017-Feb-17 at 19:17

            You can use capturing parentheses to grab the string you want:

            Source https://stackoverflow.com/questions/42305659

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install exu

            You can download it from GitHub.
            Rust is installed and managed by the rustup tool. Rust has a 6-week rapid release process and supports a great number of platforms, so there are many builds of Rust available at any time. Please refer rust-lang.org for more information.

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