beejs

 by   Ptico JavaScript Version: Current License: MIT

kandi X-RAY | beejs Summary

kandi X-RAY | beejs Summary

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

beejs
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            kandi-support Support

              beejs has a low active ecosystem.
              It has 5 star(s) with 1 fork(s). There are 1 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              beejs has no issues reported. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of beejs is current.

            kandi-Quality Quality

              beejs has no bugs reported.

            kandi-Security Security

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

            kandi-License License

              beejs 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

              beejs releases are not available. You will need to build from source code and install.
              Installation instructions are not available. Examples and code snippets are available.

            Top functions reviewed by kandi - BETA

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            beejs Key Features

            No Key Features are available at this moment for beejs.

            beejs Examples and Code Snippets

            No Code Snippets are available at this moment for beejs.

            Community Discussions

            QUESTION

            what is the content of an initialised pointer?
            Asked 2021-Apr-22 at 16:13

            I am learning about C, specifically the structs. The following exaple of a recursive (is it?) struct being used for creating a linked list. This is all from beejs guide to c.

            ...

            ANSWER

            Answered 2021-Apr-22 at 16:13

            Part 0

            The following exaple of a recursive (is it?) [...]

            (No, it isn't.)

            Part 1

            1. head = malloc(sizeof(struct node));
            2. head->data = 11;
            3. head->next = malloc(sizeof(struct node));

            on line 14 since Head is a pointer it saves the memory address from pointer. on line 16 next is the pointer but rather then using head.next to store the memory address it uses head->next. This confuses me as head -> next == (*head).next this means the content of next is assigned the memory address but next is not pointing to anything.

            head->next is equivalent to (*head).next. Assuming all the malloc calls are successful, head->next is not pointing to anything valid before the assignment, but will point to the block allocated by malloc after the assignment. For access to the member head->next (equivalently (*head).next), the important part is that the pointer value in head is valid. The probably invalid initial pointer value in head->next does not matter because it is being replaced by the assignment anyway.

            Part 2

            Source https://stackoverflow.com/questions/67215790

            QUESTION

            How to stop listening and recv data from only one user and start listening again when he disconnects
            Asked 2021-Jan-06 at 04:51

            I am trying to write a server-client program. The idea is that server

            1. listens() on the given port
            2. When a user connects it accepts() the connection and stops listening
            3. On user disconnects, it goes back to listening state and this goes on forever.

            Now, I have created the server and the communication is going fine, however, I am not sure how to stop listening when the user connects and start listening on disconnects. Can anyone help me with it?

            Also, I am following beejs guide

            Thanks

            ...

            ANSWER

            Answered 2021-Jan-05 at 04:33

            The (only) way to stop listening is to close the listening socket. That will have no effect on the already accepted connection(s), so they can continue to be used. To start listening again, you will need to open a new listening socket and bind it. You'll probably need the SO_REUSEADDR option on the socket if you want to reopen the port before the TCP delay period has elapsed.

            You could instead keep the listening socket around and simply not accept any more connections until you are finished with the first one, but that would not actually not listen -- any additional client that tried to connect would get a handshake from the kernel (so it would think it was connected) rather than a rejection.

            A third possibility which is even more unlke what you are asking, but it probably a better design would be to keep the socket open and accept additional connections while you are processing the first connection, but then close these new connections right away with some kind of BUSY message. Then clients could know at least something about what is going on.

            It all depends on what you want to have clients see when they try to connect to a busy server.

            Source https://stackoverflow.com/questions/65572989

            QUESTION

            Socket address showing 0.0.0.0, causing failure to send message
            Asked 2018-Jul-16 at 07:17

            I've been working from Beejs Network examples, introducing a few customizations. In particular, I'm trying to use a single structure to store the necessary information related to communications/sockets. I think I'm having trouble populating an addrinfo structure and using it with sendto for a UDP socket. Bellow is my code, which compiles fine, but it fails with the message outlined below

            ...

            ANSWER

            Answered 2018-Jul-16 at 07:17

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

            Vulnerabilities

            No vulnerabilities reported

            Install beejs

            You can download it from GitHub.

            Support

            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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          • HTTPS

            https://github.com/Ptico/beejs.git

          • CLI

            gh repo clone Ptico/beejs

          • sshUrl

            git@github.com:Ptico/beejs.git

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