avogadro | Avogadro 1 is not under active development | Genomics library

by cryos C++ Version: 1.2.0 License: GPL-2.0

X-Ray Key Features Code Snippets Community Discussions(3)Vulnerabilities Install Support

kandi X-RAY | avogadro Summary

avogadro is a C++ library typically used in Artificial Intelligence, Genomics applications. avogadro has no bugs, it has no vulnerabilities, it has a Strong Copyleft License and it has low support. You can download it from GitHub.

Avogadro is an advanced molecular editor designed for cross-platform use in computational chemistry, molecular modeling, bioinformatics, materials science, and related areas. It offers flexible rendering and a powerful plugin architecture. For more information, see Avogadro is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version. Avogadro is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. See INSTALL file for installation instructions.

Support

Quality

Security

License

Reuse

Support

avogadro has a low active ecosystem.

It has 322 star(s) with 154 fork(s). There are 30 watchers for this library.

It had no major release in the last 12 months.

There are 227 open issues and 648 have been closed. On average issues are closed in 178 days. There are 11 open pull requests and 0 closed requests.

It has a neutral sentiment in the developer community.

The latest version of avogadro is 1.2.0

Quality

avogadro has 0 bugs and 0 code smells.

Security

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

avogadro code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

avogadro 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.

Reuse

avogadro releases are available to install and integrate.

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avogadro Key Features

No Key Features are available at this moment for avogadro.

avogadro Examples and Code Snippets

No Code Snippets are available at this moment for avogadro.

Community Discussions

Trending Discussions on avogadro

PyLint wants a variable to be named as a constant [C0103]

How to set up and dereference a generic pointer for a derived class object for an expression parser?

Can someone tell me why I get this error message? inline constexpr variable

QUESTION

PyLint wants a variable to be named as a constant [C0103]

Asked 2022-Jan-09 at 15:56

Why does PyLint require a variable to have UPPER_CASE naming as though it is a constant?

...

ANSWER

Answered 2022-Jan-09 at 15:56

See 2 solutions here:

This is not a false positive. Pylint expects all variables at the module level to be upper case. This behaviour can be configured by passing an updated const-rgx setting in the configuration file.

Just want to add that another solution is to add logger to good-names setting in the configuration file.

Source https://stackoverflow.com/questions/70642815

QUESTION

How to set up and dereference a generic pointer for a derived class object for an expression parser?

Asked 2020-Dec-01 at 05:44

I could not find a solution to the following problem - I have a base class

...

ANSWER

Answered 2020-Dec-01 at 05:44

Is there a way to access objects of the derived class using pointers of type base class?

Yes, it is called a downcast. You usually want to avoid it since it needs you to store type information about each object, which defeats the purpose of upcasting to the base class in the first place. It can be done by dynamic_cast.

What is the proper way to construct a tree with leaves of different types and be able to traverse it using templated classes ?

Using inheritance is not the best way to approach this. Inheritance is best used when different classes need to interact with the same interface. In this case, you could use an std::variant (C++17).

Source https://stackoverflow.com/questions/65084116

QUESTION

Can someone tell me why I get this error message? inline constexpr variable

Asked 2020-Apr-18 at 20:44

I follow a tutorial from here https://www.learncpp.com/cpp-tutorial/global-constants-and-inline-variables/

main.cpp

...

ANSWER

Answered 2020-Apr-18 at 20:44

Inline variables are allowed starting from C++17.

You need to specify the -std=c++17 option on the compiler command line.

( https://en.cppreference.com/w/cpp/language/inline )

Source https://stackoverflow.com/questions/61295935

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

Vulnerabilities

No vulnerabilities reported

Install avogadro

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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