Whenever you define an IO method, you need to make sure that all IO methods
you call within it also use that argument. Change println("x ...") to
println(io, "x …").
On Friday, November 6, 2015 at 8:17:56 AM UTC-5, Gnimuc Key wrote:
>
> we can define a good representation of a new type by overloading
> `Base.show(io::IO, x::Foo)`:
>
> ```
>
> import Base: show
>
>
> type Foo
> x
> y
> end
>
> function show(io::IO, x::Foo)
> println("x ---> $(x.x)")
> println("y ---> $(x.y)")
> end
>
> julia> Foo(1,2)
> x ---> 1
> y ---> 2
>
> ```
>
> but I don't know why julia will show `Foo` 3 times when running
> `[Foo(1,2)]`:
>
> ```
>
> julia> [Foo(1,2)]
> 1-element Array{Foo,1}:
> x ---> 1
> y ---> 2
> x ---> 1
> y ---> 2
> x ---> 1
> y ---> 2
>
> ```
>
> BTW, the whitespace in the beginning of 3rd line is so wired...
>
>
