constants defeat type inference of parametric types

[stevengj]

This was unexpected:

using Base.Test
const FOO=1
type Foo{K}; end
foo() = Foo{FOO}()
@inferred foo()

gives

ERROR: return type Foo{1} does not match inferred return type Foo{K}
 in error at error.jl:19

Is it reasonable to expect Julia to be able to recognize constants in type parameters here?

cc: @simonster

[simonster]

Maybe a dup of #5560? Although in this case, we don't even need to run type inference before performing the substitution.

[stevengj]

Yah, mostly a dup.