However it can't work in such a case:
module X
export f
g(x) = 2x
function f(x)
(current_module()).g(x)
end
end
module Y
using X
export h
g(x) = 4x
h(x) = f(x)
end
module Z
using Y
println(h(4))
end
On Wed, Feb 17, 2016 at 5:37 PM, Julia Tylors <[email protected]> wrote:
> Yes,
> this worked too, Thank you.!
>
> julia> module X
> export f
> g(x) = 2x
> function f(x)
> (current_module()).g(x)
> end
> end
> X
>
> julia> module Y
> using X
> g(x) = 4x
> println(f(4))
> end
> 16
> Y
>
> julia> module Z
> using X
> g(x) = 5x
> println(f(4))
> end
> 20
> Z
>
> On Wed, Feb 17, 2016 at 4:44 PM, Lutfullah Tomak <[email protected]>
> wrote:
>
>> Hi
>> Will this achieve what you want?
>>
>> module Z
>> g()=println("Z.g")
>> function h()
>> (current_module()).g()
>> end
>> end
>> module X
>> using Z
>> g()=println("X.g")
>> function f()
>> Z.h()
>> (current_module()).g()
>> end
>> end
>> module Y
>> using X
>> g()=println("Y.g")
>> X.f()
>> end
>>
>> using Y
>>
>> It prints Y.g twice. Or do you want h() to use X's g()?
>
>
>
