Effective-Modern-Cpp | 17 best practices | Reactive Programming library

by downdemo C++ Version: Current License: Apache-2.0

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kandi X-RAY | Effective-Modern-Cpp Summary

Effective-Modern-Cpp is a C++ library typically used in Programming Style, Reactive Programming applications. Effective-Modern-Cpp has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Effective Modern C++ 主要讲述了 C++11/14 新特性的用法，阐述从两方面展开，一是原理解析，二是注意事项。如 auto 的原理使用的是模板推断机制，进而介绍模板的推断机制，再由此引出 auto 在使用上应该注意的问题。不同读者在这本书中可以各取所需，如果希望了解 C++11/14 的方方面面，则可以抽丝剥茧地深究原理，如果希望短时间内快速上手 C++11/14 工程实践，则只需略读原理但牢记结论，使用时绕开易产生问题的用法即可。此为个人笔记，条款非直译，而是个人理解的更贴近作者意图的表述。.

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Support

Effective-Modern-Cpp has a low active ecosystem.

It has 205 star(s) with 47 fork(s). There are 4 watchers for this library.

It had no major release in the last 6 months.

Effective-Modern-Cpp has no issues reported. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of Effective-Modern-Cpp is current.

Quality

Effective-Modern-Cpp has no bugs reported.

Security

Effective-Modern-Cpp has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

License

Effective-Modern-Cpp is licensed under the Apache-2.0 License. This license is Permissive.

Permissive licenses have the least restrictions, and you can use them in most projects.

Reuse

Effective-Modern-Cpp releases are not available. You will need to build from source code and install.

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Effective-Modern-Cpp Key Features

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Effective-Modern-Cpp Examples and Code Snippets

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Community Discussions

Trending Discussions on Effective-Modern-Cpp

Understand the type of param in void f(const T& param)

QUESTION

Understand the type of param in void f(const T& param)

Asked 2017-Mar-14 at 17:41

Reference: Effective Modern C++ Item 4.

https://github.com/BartVandewoestyne/Effective-Modern-Cpp/blob/master/Item04_Know_how_to_view_deduced_types/runtime_output02.cpp

...

ANSWER

Answered 2017-Jan-07 at 02:34

Question> Why the real type of param is Widget const * const & instead?

I'm not really an expert but... because T is a pointer and if you write const T & (that is T const & because the rule is that const apply on the element on the left or on the element on the right in there isn't an element on the left) you are imposing that the full T type, so the pointer, is constant.

And to impose that the pointer is constant, you have to impose cont on the left of the *.

In brief: const T & is equivalent to T const &; with T that is const Widget *, T const & become const Widget * const & or, if you prefer, Widget const * const &

Source https://stackoverflow.com/questions/41517090

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

Vulnerabilities

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Install Effective-Modern-Cpp

You can download it from GitHub.

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