Julia:函数的偶数数据类型 [英] Julia: Even-number datatype for functions

问题描述

我有大约50个函数，它们应该只使用偶数.现在，我每次都用"if"检查是否输入的数字是否为零:

I have about 50 functions which should consume only even positive numbers. Right now I am checking each time with an "if" whether the number put in is zero or not:

function grof(x::Int)
    if (x % 2) == 0
        println("good")
    else
        throw("x is not an even number!!!!!!!!!!!!! Stupid programmer!")
    end
end

理想情况下，我希望有一个数据类型可以自动生成此数据，即

Ideally, I would like to have a datatype which produces this automatically, i.e.

function grof(x::EvenInt)
    println("good")  
end

但是，由于我无法理解纪录片.感谢您的帮助！

However, I am not able to produce this datatype by my own since I am unable to understand the documentary. Thanks for your help!

最佳，v.

推荐答案

在这种情况下，我不认为创建类型是必要的:我只是

I don't think creating a type is warranted in such a situation: I would simply @assert that the condition is verified at the beginning of the function(s). (Funnily enough, checking the whether a number is even is the example that was chosen in the documentation to illustrate the effect of @assert)

例如:

julia> function grof(x::Int)
           @assert iseven(x) "Stupid programmer!"
           println("good")
       end
grof (generic function with 1 method)

julia> grof(2)
good

julia> grof(3)
ERROR: AssertionError: Stupid programmer!
Stacktrace:
 [1] grof(::Int64) at ./REPL[5]:2
 [2] top-level scope at REPL[7]:1

如果您确实要创建一个强制执行此类约束的类型，则可以.做到这一点的方法是

If you really want to create a type enforcing such a constraint, it is possible. The way to do this would be to

1. 创建一个类型(可能是Number抽象类型之一的子类型；也许是Signed)
2. 定义一个内部构造函数，以确保此类类型不能包含奇数

1. create a type (possibly subtyping one of the Number abstract types; maybe Signed)
2. define an inner constructor ensuring that such a type cannot hold an odd value

一个非常简单的示例将基于以下内容:

A very simple example to build upon would be along the lines of:

# A wrapper around an even integer value
struct EvenInt
    val :: Int

    # inner constructor
    function EvenInt(val)
        @assert iseven(val)
        new(val)
    end
end

# Accessor to the value of an EvenInt
val(x::EvenInt) = x.val

# A method working only on even numbers
grof(x::EvenInt) = println("good: $(val(x)) is even")

您将这样使用:

julia> x = EvenInt(42)
EvenInt(42)

julia> grof(x)
good: 42 is even

julia> y = EvenInt(1)
ERROR: AssertionError: iseven(val)
Stacktrace:
 [1] EvenInt(::Int64) at ./REPL[1]:5
 [2] top-level scope at REPL[6]:1

，但请注意，您还不能在EvenInt上执行任何操作:您需要解开它们(在这种情况下使用val())，或者对它们进行定义操作(如果您将其大大简化，则可以轻松完成此任务)将EvenInt设为抽象数字类型之一的子类型，并遵循相关的界面.

but note that you can't do anything on EvenInts yet: you need to either unwrap them (using val() in this case), or define operations on them (a task which can be vastly simplified if you make EvenInt a subtype of one of the abstract number types and follow the relevant interface).

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