朱莉娅:OOP 与否 [英] Julia: OOP or not

问题描述

我正在和 Julia 一起开发 Juno.

I'm working on Juno with Julia.

我不知道 Julia 是否支持 OOP.

I don't know if Julia supports OOP or not.

例如，有没有类似c++的class或struct之类的东西?

For example, is there something like class or struct of c++?

如何用数据或函数等成员声明?

How to declare it with members such as a data or a function?

推荐答案

如有疑问，请阅读文档...

When in doubt, read the documentation...

https://docs.julialang.org/en/v1/manual/types/#Composite-Types-1

长话短说:

struct MyType
    a::Int64
    b::Float64
end

x = MyType(3, 4)

x.a

方法定义在类型定义之外，例如

Methods are defined outside the type definition, e.g.

function double(x::MyType)
    x.a *= 2
end

方法不存在于类型中，例如它们在 C++ 或 Python 中的行为.这使得 Julia 的关键特性之一，即多重分派，也可以与用户定义的类型一起使用，这些类型与系统定义的类型处于完全相同的级别.

Methods do not live inside the type, as they would do in C++ or Python, for example. This allows one of the key features of Julia, multiple dispatch, to work also with user-defined types, which are on exactly the same level as system-defined types.

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