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dive | A tool for exploring each layer in a docker image | Continuous Deployment library

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kandi X-RAY | dive Summary

dive is a Go library typically used in Devops, Continuous Deployment, Docker applications. dive has no bugs, it has no vulnerabilities, it has a Permissive License and it has medium support. You can download it from GitHub, GitLab.
To analyze a Docker image simply run dive with an image tag/id/digest:.

kandi-support Support

  • dive has a medium active ecosystem.
  • It has 27275 star(s) with 1037 fork(s). There are 340 watchers for this library.
  • It had no major release in the last 12 months.
  • There are 88 open issues and 140 have been closed. On average issues are closed in 53 days. There are 9 open pull requests and 0 closed requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of dive is v0.10.0

quality kandi Quality

  • dive has 0 bugs and 0 code smells.

securitySecurity

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

license License

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

buildReuse

  • dive releases are available to install and integrate.
  • Installation instructions, examples and code snippets are available.
  • It has 6956 lines of code, 370 functions and 78 files.
  • It has medium code complexity. Code complexity directly impacts maintainability of the code.
Top functions reviewed by kandi - BETA

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Currently covering the most popular Java, JavaScript and Python libraries. See a SAMPLE HERE.
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dive Key Features

docker: Docker engine (the default option)

docker-archive: A Docker Tar Archive from disk

podman: Podman engine (linux only)

dive Examples and Code Snippets

  • dive
  • Basic Features
  • Installation
  • CI Integration
  • UI Configuration
  • ShouldProcess failing in PowerShell7
  • Flag provided but not defined: -rpc
  • Split text into smaller paragraphs of a minimal length without breaking the sentences given a threshold
  • How does React update a component and its children after a state change?
  • JavaScript if/else statements are altering variables used as conditions in the same statements
  • Is TMP really faster if the recusion depth is very deep?
  • Measuring the distance between two lines using DipLib (PyDIP)
  • Language browser detection and action rewrite
  • Rust traits with constant field defined by implementation
  • How does clang check redefinitions?

dive

dive <your-image-tag>

Community Discussions

Trending Discussions on dive
  • type deduction for std::function argument types with auto adds const
  • ShouldProcess failing in PowerShell7
  • how to skip json validation for an empty array via Golang
  • Flag provided but not defined: -rpc
  • Split text into smaller paragraphs of a minimal length without breaking the sentences given a threshold
  • How does React update a component and its children after a state change?
  • JavaScript if/else statements are altering variables used as conditions in the same statements
  • Is TMP really faster if the recusion depth is very deep?
  • Measuring the distance between two lines using DipLib (PyDIP)
  • Language browser detection and action rewrite
Trending Discussions on dive

QUESTION

type deduction for std::function argument types with auto adds const

Asked 2022-Apr-08 at 14:31

I have a struct with a method called call which has a const overload. The one and only argument is a std::function which either takes a int reference or a const int reference, depending on the overload.

The genericCall method does exactly the same thing but uses a template parameter instead of a std::function as type.

struct SomeStruct {
    int someMember = 666;

    void call(std::function<void(int&)> f) & {
        f(someMember);
        std::cout << "call: non const\n";
    }

    void call(std::function<void(const int&)> f) const& {
        f(someMember);
        std::cout << "call: const\n";
    }


    template <typename Functor>
    void genericCall(Functor f) & {
        f(someMember);
        std::cout << "genericCall: non const\n";
    }

    template <typename Functor>
    void genericCall(Functor f) const& {
        f(someMember);
        std::cout << "genericCall: const\n";
    }
};

When I now create this struct and call call with a lambda and auto & as argument the std::function always deduces a const int & despite the object not being const.

The genericCall on the other hand deduces the argument correctly as int & inside the lamdba.

SomeStruct some;
some.call([](auto& i) {
    i++;  // ?? why does auto deduce it as const int & ??
});
some.genericCall([](auto& i) {
    i++;  // auto deduces it correctly as int &
});

I have no the slightest clue why auto behaves in those two cases differently or why std::function seems to prefer to make the argument const here. This causes a compile error despite the correct method is called. When I change the argument from auto & to int & everything works fine again.

some.call([](int& i) {
    i++; 
});

When I do the same call with a const version of the struct everything is deduced as expected. Both call and genericCall deduce a const int & here.

const SomeStruct constSome;
constSome.call([](auto& i) {
    // auto deduces correctly const int & and therefore it should
    // not compile
    i++;
});
constSome.genericCall([](auto& i) {
    // auto deduces correctly const int & and therefore it should
    // not compile
    i++;
});

If someone could shine some light on this I would be very grateful!

For the more curious ones who want to dive even deeper, this problem arose in the pull request: https://github.com/eclipse-iceoryx/iceoryx/pull/1324 while implementing a functional interface for an expected implementation.

ANSWER

Answered 2022-Apr-08 at 13:25

The problem is that generic lambdas (auto param) are equivalent to a callable object whose operator() is templated. This means that the actual type of the lambda argument is not contained in the lambda, and only deduced when the lambda is invoked.

However in your case, by having specific std::function arguments, you force a conversion to a concrete type before the lambda is invoked, so there is no way to deduce the auto type from anything. There is no SFINAE in a non-template context.

With no specific argument type, both your call are valid overloads. Actually any std::function that can match an [](auto&) is valid. Now the only rule is probably that the most cv-qualified overload wins. You can try with a volatile float& and you will see it will still choose that. Once it choose this overload, the compilation will fail when trying to invoke.

Source https://stackoverflow.com/questions/71797023

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

Vulnerabilities

No vulnerabilities reported

Install dive

Available as dive in the Arch User Repository (AUR). The above example assumes yay as the tool for installing AUR packages.

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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