SimpleTraits error in Julia 0.6

General Usage
1

Is SimpleTraits still supported in 0.6? This simple code from the test suite fails. Note that it works fine on Julia 0.7.

julia> using SimpleTraits

julia> @traitdef Tr1{X}

julia> @traitimpl Tr1{Integer}

julia> const trait = SimpleTraits.trait
trait (generic function with 11 methods)

julia> trait(Tr1{Int})
Tr1{Int64}

julia> @traitfn f(x::X) where {X; Tr1{X}} = 1  # def 1
ERROR: MethodError: no method matching start(::Void)
Closest candidates are:
  start(::SimpleVector) at essentials.jl:258
  start(::Base.MethodList) at reflection.jl:560
  start(::ExponentialBackOff) at error.jl:107
  ...
2

Sorry, but the where function syntax is not supported in Julia 0.6 due to the {X; Tr{X}} not parsing as needed: where-syntax accepts ; as separator in 0.6 · Issue #23211 · JuliaLang/julia · GitHub. Note though that the old function syntax still works in 0.7, but of course throws deprecation warnings.

3

Thanks, I have found examples from the old README file.

@traitfn f{X; Tr1{X}}(x::X) = 1
4

Yep, this was pretty hidden. I now added a link back to the last 0.6 version to the README: https://github.com/mauro3/SimpleTraits.jl/tree/master#manual.