Question

我想用参数类型Family创建类型Family，但我想用Null参数创建可能性

type Family
    name:: AbstracDtring
    people:: Int
    dad:: Family
    mom:: Family

Family(name:: AbstractString, people::Int ) = new (name, people , NULL, NULL)


end

我可以做这个事情我想创建“对象”引用另一个对象或没有引用

Answer 1

您可以使用较少的参数调用new：

type Family
    name::AbstractString
    people::Int
    dad::Family
    mom::Family

    Family(name::AbstractString, people::Int) = new(name, people)
end

您可以构建实例，但在分配.dad和.mom字段之前，访问它们会导致错误：

julia> fam = Family("Jones", 3)
Family("Jones",3,#undef,#undef)

julia> fam.dad
ERROR: UndefRefError: access to undefined reference
 in eval(::Module, ::Any) at ./boot.jl:225
 in macro expansion at ./REPL.jl:92 [inlined]
 in (::Base.REPL.##1#2{Base.REPL.REPLBackend})() at ./event.jl:46

julia> fam.mom
ERROR: UndefRefError: access to undefined reference
 in eval(::Module, ::Any) at ./boot.jl:225
 in macro expansion at ./REPL.jl:92 [inlined]
 in (::Base.REPL.##1#2{Base.REPL.REPLBackend})() at ./event.jl:46

julia> fam.dad = fam
Family("Jones",3,Family(#= circular reference @-1 =#),#undef)

julia> fam.mom
ERROR: UndefRefError: access to undefined reference
 in eval(::Module, ::Any) at ./boot.jl:225
 in macro expansion at ./REPL.jl:92 [inlined]
 in (::Base.REPL.##1#2{Base.REPL.REPLBackend})() at ./event.jl:46

您可以使用isdefined功能检查字段是否已定义：

julia> isdefined(fam, :dad)
true

julia> isdefined(fam, :mom)
false

Nullable方法也有效，但重量稍轻一些。

Answer 2

使用Nullable（http://docs.julialang.org/en/release-0.4/manual/types/#nullable-types-representing-missing-values）

type Family
    name:: AbstractString
    people:: Int
    dad:: Nullable{Family}
    mom:: Nullable{Family}

Family(name:: AbstractString, people::Int ) = new(name, people, Nullable{Family}(), Nullable{Family}())

end

如何在Julia中使用Null argumnet创建构造函数

2 个答案: