[英]Recover type in parametric composite type
In Julia (< 0.6), when creating a parametric composite type such as MyType{T}
, is there a clean way to recover T
from an instance of that type? 在Julia(<0.6)中,当创建参数复合类型(如
MyType{T}
,是否有一种从该类型的实例中恢复T
的简洁方法?
Take their example from the doc: 以他们的文档为例:
type Point{T}
x::T
y::T
end
I can create an object p = Point(5.0,5.0)
, T
here will be matched to Float64
so that the corresponding object is a Point{Float64}
. 我可以创建一个对象
p = Point(5.0,5.0)
,这里T
将与Float64
匹配,以便相应的对象是Point{Float64}
。 Is there a clean way to recover Float64
here? 有一个干净的方法来恢复
Float64
吗?
I could do 我可以做
typeof(p.x)
But it feels like that's not the right thing to do. 但感觉这不是正确的事情。
When you need the type parameter, you should define a parametric method. 如果需要type参数,则应定义参数方法。 That is the only proper way to access the type parameter.
这是访问type参数的唯一正确方法。
So for a Point
, 所以对于一个
Point
,
function doSomething{T}(p::Point{T})
// You have recovered T
println(T)
end
The type is saved in the class information: 类型保存在类信息中:
typeof(Point(1, 2)).parameters # -> svec(Int64)
It's more general than writing a specific function for it, but I'm not sure it's considered official. 它比为它编写特定功能更为通用,但我不确定它是否被认为是正式的。
There's also fieldtype
还有
fieldtype
fieldtype(typeof(Point(1.0, 1.0)), :x) # --> Float64
fieldtype(Point, :x) # --> T
fieldtype(Point{Int64}, :x) # --> Int64
Not sure how that's any better than just getting the type of the instance though. 不知道如何才能获得实例的类型。
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