Here the compiler already knows that foo returns its input
reference. So it checks whether foo is being passed a local -
no; but it also has to check if foo is passed any ref
parameters of quux, which it is. The compiler now has to mark
quux as a function that returns its input reference.
Works?
If functions's source isn't available, the compiler can't know
what the function does.
This could only work if this property of a function (whether it
returns a reference to its ref parameter) would be part of its
type. The compiler could still infer it for function literals and
templates, similar to how pure works now.
