Is it possible to define a default argument value based on the function's internal variables at the moment of return?

Internals & Design
1

Would it possible and appropriate for Julia to extend functionality of function definitions to include ones like this?

function f(x, z = y^2)
    y = 2x
    return x + y + z
end

It would need to evaluate the default value of z as the (second) last command for the function f since the value of its dependent variable y could get changed anytime during the function call.

Is it possible and appropriate to design such functionality? I feel like there may be potential clashes somehow (which I haven’t been able to think of yet, but maybe something like default interdependent variables e.g. function g(x, z = 2y, y = 3z) ... end) but those could be parsed as errors.

2 
function f(x, z = nothing)
   y = 2x
   tmp = z === nothing ? y^2 : z
   return x + y + tmp
end

Because y only exists in the function body, it can’t be used in the default arguments. Default arguments can depend on other arguments though:

julia> f(a, b=2a) = b
f (generic function with 2 methods)

julia> f(1)
2
6 Likes
4

Also, if you are OK with keyword arguments and y visible to the user, you can do

f(x; y = 2*x, z = abs2(y)) = x + y + z
f(x)
f(x; z = 3)
5 Likes