hexchat | IRC client for Windows and UNIX-like operating systems | Chat library

 by   hexchat C Version: v2.16.1 License: GPL-2.0

kandi X-RAY | hexchat Summary

kandi X-RAY | hexchat Summary

hexchat is a C library typically used in Messaging, Chat applications. hexchat has no bugs, it has no vulnerabilities, it has a Strong Copyleft License and it has medium support. You can download it from GitHub, GitLab.

HexChat is an IRC client for Windows and UNIX-like operating systems. See IRCHelp.org for information about IRC in general. For more information on HexChat please read our documentation:.
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              hexchat has a medium active ecosystem.
              It has 2867 star(s) with 513 fork(s). There are 113 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 736 open issues and 1387 have been closed. On average issues are closed in 452 days. There are 33 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of hexchat is v2.16.1

            kandi-Quality Quality

              hexchat has 0 bugs and 0 code smells.

            kandi-Security Security

              hexchat has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              hexchat code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              hexchat is licensed under the GPL-2.0 License. This license is Strong Copyleft.
              Strong Copyleft licenses enforce sharing, and you can use them when creating open source projects.

            kandi-Reuse Reuse

              hexchat releases are available to install and integrate.

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            hexchat Key Features

            No Key Features are available at this moment for hexchat.

            hexchat Examples and Code Snippets

            No Code Snippets are available at this moment for hexchat.

            Community Discussions

            QUESTION

            Does comparing a pointer that has been free'd invoke UB?
            Asked 2019-Sep-17 at 22:30

            This seems to be a fairly common pattern e.g. in hexchat (may not compile, see also plugin docs. also note that hexchat_plugin_get_info hasn't been used in forever so I'm omitting it for simplicity):

            ...

            ANSWER

            Answered 2018-Oct-03 at 14:46

            Using a value of a pointer after the object it is pointing to have reached it's lifetime end is indeterminate as stated in the C11 Standard draft 6.2.4p2 (Storage durations of objects) (the emphasis is mine):

            The lifetime of an object is the portion of program execution during which storage is guaranteed to be reserved for it. An object exists, has a constant address, and retains its last-stored value throughout its lifetime. If an object is referred to outside of its lifetime, the behavior is undefined. The value of a pointer becomes indeterminate when the object it points to (or just past) reaches the end of its lifetime.

            And using it's value (just for anything) is an explicit undefined behavior as stated in Annex J.2(Undefined behavior):

            The behavior is undefined in the following circumstances: [...] The value of a pointer to an object whose lifetime has ended is used (6.2.4).

            Source https://stackoverflow.com/questions/52628773

            QUESTION

            What is the proper way to terminate a Timer thread in python?
            Asked 2019-Aug-13 at 08:50

            The only other question I can find about terminating Python timer threads is this one: how to terminate a timer thread in python but that only tells me what I already knew from reading the threading documentation.

            I am writing a GTK app in Python. It is interfacing with HexChat via d-bus. Because there is no D-bus signal when a user changes context (switches to a different IRC channel or server window) my original design waited until the user interacted with the GUI in some way to see what context we are currently in. The downside to this is entry boxes which need to be switched out can't really be used to switch contexts, because if a user starts typing, the changed signal is issued and the contents get saved to the old context, which is incorrect behaviour.

            To solve this, I decided to use a threaded timer task to check what the current context is every 0.5 seconds. The thread function checks the current context, updates a variable in the class, and then re-starts itself with another 0.5 second delay. This works perfectly and is fast enough that it will keep up with users switching channels often.

            However, even when I add a TimerThread.cancel to my __del__() function, instead of my program exiting it just hangs until I give a keyboard interrupt. I also tried doing a TimerThread.cancel, sleep(0.1), TimerThread.cancel again just in case I happened to cancel the timer as it was in the context-checking function. Same result. I also tried putting the TimerThread.cancel in the onDestroy function (which will in turn call the __del__ destructor) but no change. The GTK loop exits, and the GUI disappears, but the program just hangs in the console until keyboard interrupt. When I do give a keyboard interrupt, I get a traceback error from the threading library.

            Is there some other way to terminate the thread? Is there some other step I need to take to kill the threading library before exiting? Am I misunderstanding how Timer threads work?

            Here are the relevant sections of the code: https://paste.ubuntu.com/p/kQVfF78H5R/

            EDIT: Here is the traceback, if that helps.

            ...

            ANSWER

            Answered 2019-Aug-13 at 08:50

            I think you're running into a race condition. Try to protect the contextUpdater variable with a lock. Here, I added three new functions, namely start_updater which is called from the class __init__ method, stop_updater, and restart_updater which is called from the getContextTimer method. Before exiting your application, call stop_updater method which will cancel the currently running timer (don't call it from __del__ method):

            Source https://stackoverflow.com/questions/57470601

            QUESTION

            InspIRCd Secure WebSocket Connections Fail
            Asked 2019-Jun-27 at 21:46

            When attempting to connect to a socket configured with SSL and the WebSocket Hook the connection fails.

            JavaScript WebSocket Request:

            ...

            ANSWER

            Answered 2017-Jul-12 at 16:50

            You have to accept Self-Signing cert on https://ohmingle.com:7001 first and you can connect via wss.

            Source https://stackoverflow.com/questions/45063211

            QUESTION

            How to remove duplicates from list of dicts?
            Asked 2019-Mar-24 at 14:44

            I have a list of dictionaries in python as follows:

            ...

            ANSWER

            Answered 2019-Mar-23 at 19:06

            Use a set to keep track of all (category, name) pairs you've already seen:

            Source https://stackoverflow.com/questions/55317251

            QUESTION

            How to use fsockopen (or compatible) with SOCKS proxies in PHP?
            Asked 2019-Mar-13 at 21:08

            I've coded a non-evil, non-spammy IRC bot in PHP, using fsockopen and related functions. It works. However, the problem is that I need to support proxies (preferably SOCKS5, but HTTP is also OK if that is somehow easier, which I doubt). This is not supported by fsockopen.

            I've gone through all search results for "PHP fsockopen proxy" and related queries. I know of all the things that don't work, so please don't link to one of them.

            The PHP manual page for fsockopen mentions the function stream_socket_client() as

            similar but provides a richer set of options, including non-blocking connection and the ability to provide a stream context.

            This sounded promising at first, supposedly allowing me to just replace the fsockopen call with stream_socket_client and specify a proxy, maybe via a "stream context"... but it doesn't. Or does it? I'm very confused by the manual.

            Please note that it must be a PHP code solution; I cannot pay for "Proxifier" or use any other external software to "wrap around" this.

            All the things I've tried seem to always result in me getting a bunch of empty output from the server, and then the socket is forcefully closed. Note that the proxy I'm trying with works when I use HexChat (a normal IRC client), with the same network, so it's not the proxies themselves that are at fault.

            ...

            ANSWER

            Answered 2019-Mar-13 at 21:08

            As far as I know there is no default option to set a SOCKS or HTTP proxy for fsockopen or stream_socket_client (we could create a context and set a proxy in HTTP options, but that doesn't apply to stream_socket_client). However we can establish a connection manually.

            Connecting to HTTP proxies is quite simple:

            • The client connects to the proxy server and submits a CONNECT request.
            • The server responds 200 if the request is accepted.
            • The server then proxies all requests between the client and destination host.

            Source https://stackoverflow.com/questions/54863701

            QUESTION

            NickServ HexChat Client Locked Username
            Asked 2017-Feb-17 at 21:16

            I am using the HexChat client on my local machine. I have been using an unregistered nick, eg. foobar. When my connection has dropped I get a message when I try to reconnect:

            foobar is already in use. Retrying with foobar01... foobar01 is already in use. Retrying with foobar02...

            This has been like this for weeks. My actual username I am trying to use is not something anyone wold be using so it seems it is locked. Obviously I cannot register the nick because I can't connect with it.

            What's the solution?

            ...

            ANSWER

            Answered 2017-Feb-17 at 21:16

            If you are saying it is happening only temporarily after disconnecting that is often referred to as a ghost and your old connection just hasn't timed out. If you register the nick you can forcefully disconnect it otherwise you wait.

            If it is always happening and another user is actually using it /whois $thenick then there is nothing you can do about that since you didn't register it.

            Source https://stackoverflow.com/questions/42294099

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

            Vulnerabilities

            No vulnerabilities reported

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            You can download it from GitHub, GitLab.

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            https://github.com/hexchat/hexchat.git

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            gh repo clone hexchat/hexchat

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            git@github.com:hexchat/hexchat.git

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