::Type{T}=Float64 argument breaks hygiene?

[phipsgabler]

If this is not a misinterpretation on my side, I find the following a bit shocking:

julia> macro test1()
           return quote
               function foo(::Type{T}=Float64) where {T}
                   return Vector{T}(undef, 10)
               end
           end
       end
@test1 (macro with 1 method)

julia> @test1()
ERROR: syntax: malformed expression
Stacktrace:
 [1] top-level scope
   @ REPL[3]:1

julia> function foo(::Type{T}=Float64) where {T}
           return Vector{T}(undef, 10)
       end
foo (generic function with 2 methods)

julia> foo()
10-element Vector{Float64}:
 0.0
 0.0
 0.0
 0.0
 0.0
 0.0
 0.0
 0.0
 0.0
 0.0

julia> versioninfo()
Julia Version 1.6.2
Commit 1b93d53fc4 (2021-07-14 15:36 UTC)
Platform Info:
  OS: Linux (x86_64-pc-linux-gnu)
  CPU: Intel(R) Core(TM) i5-6200U CPU @ 2.30GHz
  WORD_SIZE: 64
  LIBM: libopenlibm
  LLVM: libLLVM-11.0.1 (ORCJIT, skylake)

Over on Zulip, someone suggested to uncomment some debug printing in jlfrontend.scm -- the top of the Lisp stack trace you get is (for a different example macro with variable TV, though):

(type-error car cons ())
unexpected error: #0 (caddr (:: (curly Type TV)))
#1 (resolve-expansion-vars-
 (kw (:: (curly Type TV)) (curly Vector Float64))
 ((n . |#118#n|) (|#unused#| . |#119##unused#|) (TV . |#120#TV|))
 #<julia: DynamicPPL> () #t)

But I'm unable to interpret that. (Here's the full example, run on the phg/unescape branch of Turing.jl.)