Definition of function with keyword arguments via macro does not work as expected

[jonas-hagen]

Consider the following example:

macro myDef1(name)
    return esc(:( $name(x; y=3) = x*y ))
end
@myDef1 f1
f1(4) # gives 3 but 12 expected
f1(4, y=5) # gives 5 but 20 expected

Or:

macro myDef2(name)
    return esc(:( $name(x; y=3) = x )) # not using y here!
end
@myDef2 f2 # gives ERROR: syntax: malformed expression

Whereas if the return statment within the function is expicitly written, it works:

macro myDef3(name)
    return esc(:( $name(x; y=3) = return (x*y) ))
end
@myDef3 f3
f3(4) # now gives 12, as expected
f3(4, y=5) # now gives 20, as expected

Julia Version 0.3.0
Commit 7681878 (2014-08-20 20:43 UTC)

[eschnett]

Keyword arguments look like assignments, but they are not -- they are :kw expressions:

julia> dump(:(f(a=2,b)))
Expr 
  head: Symbol call
  args: Array(Any,(3,))
    1: Symbol f
    2: Expr 
      head: Symbol kw
      args: Array(Any,(2,))
        1: Symbol a
        2: Int64 2
      typ: Any
    3: Symbol b
  typ: Any

I assume that, in the first case, the parser treats the content of esc(...) expression as assignment statement, so that its LHS (left hand side) is a sequence of statements ("x" and "y=3"), where "y=3" is an assignment again. In your second case, the return statement forces the parser to treat this as function definition, and "y=3" is then correctly parsed as keyword argument.

If this is true, then using the "function" syntax for your function definition should help as well.

[JeffBezanson]

cc @JuliaBackports