Re: [julia-users] How to generate composite types with a macro?

[Kaj Wiik]

Thanks for the reply!

Hmm, well my intention was to build the definition piecewise inserting the 
fields after this line without end, now I see that it will not work. 
Perhaps I should make a template and push! fields to ex.args[3].args?

In fact your suggestion does not work in my Julia:
julia> macro deftype(name)
        ex1 = :(immutable $name end)
       end

julia> @deftype foo

julia> foo()
ERROR: foo not defined



The reason is:
julia> macroexpand(:(@deftype foo))
:(immutable #133#foo
    end)


I.e. the type name is modified... How to escape it?

Thanks,
Kaj

julia> versioninfo()
Julia Version 0.3.6
Commit a05f87b* (2015-01-08 22:33 UTC)
Platform Info:
  System: Linux (x86_64-linux-gnu)
  CPU: AMD A6-4455M APU with Radeon(tm) HD Graphics
  WORD_SIZE: 64
  BLAS: libopenblas (NO_LAPACK NO_LAPACKE DYNAMIC_ARCH NO_AFFINITY)
  LAPACK: liblapack.so.3
  LIBM: libopenlibm
  LLVM: libLLVM-3.3


On Friday, March 6, 2015 at 5:35:23 PM UTC+2, Isaiah wrote:
>
> Missing an `end` in both expressions.
>
>>
> julia> macro deftype(name)
>
>>            ex1 = :(immutable $name end)
>          end
> julia> @deftype foo
> julia> foo()
>
> On Thu, Mar 5, 2015 at 5:02 AM, Kaj Wiik <kaj....@gmail.com <javascript:>> 
> wrote:
>
>> I have been trying to write a macro that would generate fields in a 
>> for-loop.
>>
>> However, when I try to generate the first line I get an error:
>>
>> julia> macro deftype(name,artype,num)
>>        ex1 = :(esc( immutable $name))
>> ERROR: syntax: unexpected ")"
>>
>> julia> macro deftype(name,artype,num)
>>        ex1 = :(esc(quote immutable $name end))
>> ERROR: syntax: extra token ")" after end of expression
>>
>>
>> There must be a way to do this, I cannot find how...
>>
>> Thanks,
>> Kaj
>>
>>
>