fieldtype returns too-general type

[eschnett]

In the function below, Julia returns Any for the field type of a tuple, although Julia actually knows that the field has type Int:

function f{T}(::T)
    @show T
    @show [fieldtype(T, i) for i in 1:nfields(T)]
    nothing
end
f(((nothing, nothing), (nothing, 4)))

I see the output:

T = Tuple{Tuple{Void,Void},Tuple{Void,Int64}}
[fieldtype(T,i) for i = 1:nfields(T)] = DataType[Tuple{Void,Any},Tuple{Void,Any}]

This is different from Julia 0.5, which returns the fully specific type:

T = Tuple{Tuple{Void,Void},Tuple{Void,Int64}}
[fieldtype(T,i) for i = 1:nfields(T)] = DataType[Tuple{Void,Void},Tuple{Void,Int64}]

#12793 might be related.

[vtjnash]

backport is pending #17953

[eschnett]

@vtjnash That went the other way around than I expected. I expected 0.6 to return the more specific type, not 0.5 to return a less specific type.

Put differently: How can I access the specific type information that is output by @show?

[vtjnash]

ok, I read that backwards: probably /broken/ by #17953

[eschnett]

Here is a work-around (it's not pretty):

@generated function ft{T, i}(x::T, ::Type{Val{i}})
    :($(fieldtype(x, i)))
end

function f{T}(x::T)
    @show T
    @show [fieldtype(T, i) for i in 1:nfields(T)]
    @show [ft(x, Val{i}) for i in 1:nfields(T)]
    nothing
end

f(((nothing, nothing), (nothing, 4)))

This outputs

T = Tuple{Tuple{Void,Void},Tuple{Void,Int64}}
[fieldtype(T,i) for i = 1:nfields(T)] = DataType[Tuple{Void,Any},Tuple{Void,Any}]
[ft(x,Val{i}) for i = 1:nfields(T)] = DataType[Tuple{Void,Void},Tuple{Void,Int64}]

[vtjnash]

yes, the @generated ensures that the compiler can't optimize it