Workaround UndefVarError: A not defined
#156
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The above error happens when I do using ForwardDiff2: D; using ModelingToolkit
@variables a[1:3]
D(D(prod))(a)' |
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I would like to understand this a bit better first, which of the two types are undefined? This needs an upstream bug-report as well. The failing tests are #155 |
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With diff --git a/src/context.jl b/src/context.jl
index d02a16d..d98b74f 100644
--- a/src/context.jl
+++ b/src/context.jl
@@ -451,7 +451,9 @@ See also: [`canrecurse`](@ref), [`overdub`](@ref), [`recurse`](@ref), [`prehook
# to `Core.apply_type`. In the future, it would be best for Julia's compiler to better handle
# varargs calls to such functions with type arguments, or at least provide a better way to
# force specialization on the type arguments.
-@inline call(::ContextUntagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = @isdefined(A) && @isdefined(B) ? f(A, B) : f(a, b)
+@inline function call(::ContextUntagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B}
+ @isdefined(A) && @isdefined(B) ? f(A, B) : (Main.__a[] = a; f(@show(a), b))
+end
@inline call(::ContextTagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = @isdefined(A) && @isdefined(B) ? f(A, B) : f(a, b)
"""I got julia> using ForwardDiff2: D; using ModelingToolkit
julia> @variables v[1:3];
julia> __a = Ref{Any}();
julia> D(D(prod))(v)';
a = ForwardDiff2.Dual{ForwardDiff2.Tag{Nothing},V,P} where P
julia> dump(__a[])
UnionAll
var: TypeVar
name: Symbol P
lb: Core.TypeofBottom Union{}
ub: Any
body: ForwardDiff2.Dual{ForwardDiff2.Tag{Nothing},V,P} <: Real
value::V
partials::PThat is super weird... |
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vtjnash
reviewed
May 26, 2020
Comment on lines
+454
to
+455
| @inline call(::ContextUntagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = @isdefined(A) && @isdefined(B) ? f(A, B) : f(a, b) | ||
| @inline call(::ContextTagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = @isdefined(A) && @isdefined(B) ? f(A, B) : f(a, b) |
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Suggested change
| @inline call(::ContextUntagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = @isdefined(A) && @isdefined(B) ? f(A, B) : f(a, b) | |
| @inline call(::ContextTagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = @isdefined(A) && @isdefined(B) ? f(A, B) : f(a, b) | |
| @inline call(::ContextUntagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = f(a, b) | |
| @inline call(::ContextTagged, f::typeof(Core.apply_type), a::Type{A}, b::Type{B}) where {A,B} = f(a, b) |
where {A} preserves only the equality of a type and not its identity. Thus, especially in a context like this, you must always use the value, to avoid corrupting the meaning of the expression. For example, the former may equate Dict{K,V} where {K,V} with Dict{K,V} where {V,K} since they are equivalent types, but the call to apply_type should yield very different results for those two based upon their distinct identity.
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This PR fixes the following error, though, maybe something more sinister is going on. Unfortunately, I cannot find an MWE. I understand that this PR is not perfect. However, for the time being, I wonder if this PR can be merged and tagged since it is blocking the release of ForwardDiff2.