[julia-users] Re: how to call c function from Julia to pass struct params?

[Ivar Nesje]

If you never need to look at the fields of the struct, and the return value 
is a pointer to the struct, you can declare the return type as Ptr{Void}.

For type safety, you might want to wrap the Ptr{Void} in a Julia immutable

immutable MyType
    ptr::Ptr{Void}
end

function MyType(a::Integer, b::Integer)
    MyType(ccall( (:com1, "pathToLib"), Ptr{Void}, (Int32, Int32),a ,b))
end

function draw(a::MyType)
    MyType(ccall( (:draw, "pathToLib"), nothing, (Ptr{void}),a))
end

If you need to access the fields of the struct in Julia, you might have 
success trying StrPack.

If myType is passed around by value (as the signature of draw indicates by 
omitting the *), you have to find another solution.

Ivar

kl. 10:14:31 UTC+2 tirsdag 1. juli 2014 skrev huangb...@gmail.com følgende:
>
> I am a newbee in Julia. I want to call a c function in Julia, but I have a 
> problem.
> In C:
> function1:
>
>    myType com1(int a, int b)
>    {
>         do some operator;
>         return myType;
>    }
>
> "myType" is a struct type that I define
>
> function2:
>
>    void draw(myType a)
>   {
>       do some operator 
>       return ;
>   }
>
> In Julia 
> ccall( (:com1, "pathToLib"), ReturnedType, (Int32, Int32),a ,b)
> but "ReturnedType" should responsed to "myType", but julia do not have the 
> type, what should I do? and how to pass the return value to function2?
>
>